inequalities - florida atlantic universitymath.fau.edu/yiu/mps2016/psrm2016k.pdf · 2016. 3....
TRANSCRIPT
![Page 1: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/1.jpg)
Chapter 27
Inequalities
27.1 The inequality AGH2
Theorem 27.1. For positive numbers x and y,
x+ y
2≥ √
xy ≥ 2xy
x+ y.
Equality holds if and only if x = y.
G′ GA′ A
X′ X
Y ′ Y
O
Q
P
HH′K
Proof. Let XX ′ = x, Y Y ′ = y.
1. AA′ = 12(x+ y) = XY .
2. GG′ = diameter of circle = PQ =√(
x+y2
)2 − (x−y2
)2=
√xy.
3. HH ′ = 2xyx+y
. 1
1Proof: HKx
+ HKy
= Y HXY
+ HXXY
= Y H+HXXY
= 1; 1HK
= 1x+ 1
y. This means that 2
HH′ = 1x+ 1
y,
and HH′ = 2xyx+y
.
![Page 2: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/2.jpg)
418 Inequalities
4. AA′ ≥ GG′ ≥ HH ′. Equality holds if and only if the trapezoid isa square.
Examples
1. Prove that if a, b, c are positive numbers, then (a+b)(b+c)(c+a) ≥8abc.
Proof. (a+ b)(b+ c)(c+ a) ≥ 2√ab · 2√bc · 2√ca = 8abc.
2. (Easy) Prove that 2(x2+y2) ≥ (x+y)2. When does equality hold?
3. Let a, b, c, d be positive numbers such that a+ b+ c+d = 1. Showthat √
4a+ 1 +√4b+ 1 +
√4c+ 1 +
√4d+ 1 < 6.
Proof.√4a+ 1 =
√(4a+ 1) · 1 < 4a+1+1
2= 2a + 1; similarly
for b, c, d. Adding up the four inequalities, we obtain√4a+ 1+
√4b+ 1+
√4c+ 1+
√4d+ 1 < 2(a+b+c+d)+4 = 6.
4. For positive numbers x1, x2, . . . , xn,
x21
x2
+x22
x3
+ · · ·+ x2n−1
xn
+x2n
x1
≥ x1 + x2 + · · ·+ xn.
2
5. For positive real numbers x, y, z,
x3 + y3 + z3 ≥ 1
3(x+ y + z)(x2 + y2 + z2).
3 Generalization: 4
xn + yn + zn ≥ 1
3(x+ y + z)(xn−1 + yn−1 + zn−1).
2∑ x2i
xi+1≥ ∑
(2xi − xi+1) =∑
xi.3Chinese Mathematical Bulletin, Problem 1128, April, 1998.4Chinese Mathematical Bulletin, January, 2000, 24–25;19.
![Page 3: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/3.jpg)
27.1 The inequality AGH2 419
Equality holds if and only if n = 1 or x = y = z.
More generally,
m∑i=1
xni ≥ 1
m
( m∑i=1
xi
)( m∑i=1
xn−1i
).
m∑i=1
xni ≥ 1
m
( m∑i=1
xki
)( m∑i=1
xn−ki
).
• If x+ y + z = 1, then x2 + y2 + z2 ≥ 13.
• If∑
xi = 1, then
∑(xi +
1
xi
)2 ≥ (n2 + 1)2
n.
• (s− a)4 + (s− b)4 + (s− c)4 ≥ �2.
6. For n > 1, 1 + 1√2+ 1√
3+ · · ·+ 1√
n>
√n. 5
7. Show that n! < (n+12)n.
8. Show that 1 · 3 · 5 · · · (2n− 1) < nn.
9. For all positive integers n, show that
1
2· 34· 56· · · 2n− 1
2n≤ 1√
3n+ 1.
10. Show that
13 + 23 + · · ·+ (n− 1)3 <n4
4< 13 + 23 + · · ·+ n3.
11. Prove that f(x) = sin x is convex on [0, π], and hence
1
n(sin θ1 + sin θ2 + · · ·+ sin θn) ≤ sin
1
n(θ1 + θ2 + · · ·+ θn)
for 0 ≤ θi ≤ π.5Chinese Mathematical Bulletin, June, 2000, 28–29.
![Page 4: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/4.jpg)
420 Inequalities
12. Prove that if k is a positive integer, then√k + 1 +
√k − 1 < 2
√k.
Hence deduce that
1√k<
√k + 1−√
k − 1.
Making use of the above, or otherwise, prove that for every positiveinteger n,
1 +1√2+
1√3+ · · ·+ 1√
n<
√n+ 1 +
√n− 1.
13. Which is larger: eπ or πe?
27.2 Proofs of the A-G inequality
The Arithmetic-Geometric Means Inequality For positive real num-bers a1, a2, . . . , an, let
G(a1, a2, . . . , an) = n√a1a2 · · · an, and A(a1, a2, . . . , an) =
1
n(a1+a+ · · ·+an).
(a) Show that if a1 = a2 = · · · = an = a, then
G(a1, a2, . . . , an) = A(a1, a2, . . . , an) = a.
(b) Show that if a1, a2, . . . , an are not all equal then there is one aiwhich is greater than G(a1, a2, . . . , an), and there is one aj whichis less than G(a1, a2, . . . , an).
(c) Suppose a1 > G(a1, a2, . . . , an) > an. Let
a′1 = G(a1, a2, . . . , an),
a′2 = a2, . . . , a′n−1 = an−1, and
a′n = a1anG(a1,a2,...,an)
.
Show that G(a′1, a′2, . . . , a
′n) = G(a1, a2, . . . , an),
and A(a′1, a′2, . . . , a
′n) < A(a1, a2, . . . , an).
![Page 5: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/5.jpg)
27.2 Proofs of the A-G inequality 421
(d) Using the results above, show that
G(a1, a2, . . . , an) ≤ A(a1, a2, . . . , an),
and that equality holds if and only if a1 = a2 = · · · = an.
Cauchy - Schwarz Inequality For real numbers x1, x2, . . . , xn; y1,y2, . . . , yn,
(x1y1+x2y2+ · · ·+xnyn)2 ≤ (x2
1+x22+ · · ·+x2
n)(y21 + y22 + · · ·+ y2n).
Equality holds if and only if
x1 : y1 = x2 : y2 = · · · = xn : yn.
Inequalities from Convexity Let f(x) be a convex function definedon [a, b]:
f(x1) + f(x2) ≤ 2f(x1 + x2
2)
for all x1, x2 ∈ [a, b]. For each positive integer n, consider the statement
I(n) : For xi ∈ [a, b], 1 ≤ i ≤ n,
f(x1) + f(x2) + · · ·+ f(xn) ≤ n · f(x1 + x2 + · · ·+ xn
n).
(i) Prove by induction that I(2k) is true for every positive integer k.
(ii) Prove that if I(n), n ≥ 2, is true, then I(n− 1) is true.
(iii) Prove that I(n) is true for every positive integer n.
[H228.] Verify that the following three inequalities hold for positivereals x, y, and z:
(i) x(x− y)(x− z) + y(y − x)(y − z) + z(z − x)(z − y) ≥ 0. [Thisis known as Schur’s inequality].
(ii) x4 + y4 + z4 + xyz(x+ y + z) ≥ 2(x2y2 + y2z2 + z2x2).
(iii) 9xyz + 1 ≥ 4(xy + yz + zx), where x+ y + z = 1.Can you devise an ingenious method that allows you to solve the
problem without having to prove all three inequalities directly ?
![Page 6: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/6.jpg)
422 Inequalities
27.3 The AGH mean inequality
Theorem 27.2. Let x1, x2, . . . , xn be positive numbers.
A(x1, x2, . . . , xn) ≥ G(x1, x2, . . . , xn) ≥ H(x1, x2, . . . , xn).
Equality holds if and only if x1 = x2 = · · · = xn.
Proof. We first prove the AG mean inequality by induction on the num-ber of variables. The validity of AG2 has already been established. As-suming AGn, we consider n + 1 nonnegative numbers x1, x2, . . . , xn,xn+1, with arithmetic mean A and geometric mean G. Applying AGn tox1, x2, . . . , xn, and also to xn+1 and n− 1 copies of G, we have
x1 + x2 + · · ·+ xn ≥n n√x1x2 · · · xn,
xn+1 + (n− 1)G ≥n n√
xn+1 ·Gn−1.
Combining these two inequalities and applying AG2, we have
(n+ 1)A+ (n− 1)G =(x1 + x2 + · · ·+ xn) + (xn+1 + (n− 1)G)
≥n n√x1x2 · · · xn + n n
√xn+1 ·Gn−1
≥2n
√n√x1x2 · · · xn · n
√xn+1 ·Gn−1
=2n 2n√x1x2 · · · xnxn+1 ·Gn−1
=2n2n√Gn+1 ·Gn−1
=2n ·G.
From this, A ≥ G. Equality holds if and only if x1 = x2 = · · · = xn,and xn+1 = G = n+1
√xn1xn+1, which entails xn+1 = x1. This proves
AGn+1 and completes the proof of the AG mean inequality. The GHmean inequality now follows easily.
H(x1, x2, . . . , xn) =1
A( 1x1, 1x2, . . . , 1
xn)≤ 1
G( 1x1, 1x2, . . . , 1
xn)= G(x1, x2, . . . , xn).
There are many proofs of the AG mean inequality. A very elegantproof of a stronger inequality (by H. Alzer) appeared recently in theAmerican Mathematical Monthly. 6
6H. Alzer, A proof of the arithmetic mean - geometric mean inequality, American Mathematical Monthly,103 (1996) 585.
![Page 7: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/7.jpg)
27.3 The AGH mean inequality 423
For positive real numbers a1 ≤ a2 ≤ · · · ≤ an, and p1, p2, . . . , pnsatisfying p1 + p2 + · · ·+ pn, let
A =n∑
k=1
pkak, G =n∏
k=1
apkk .
The number G lies between some ah and ah+1. The sum
n∑k=1
pk
∫ G
ak
(1
t− 1
G
)dt =
h∑k=1
pk
∫ G
ak
(1
t− 1
G
)dt+
n∑k=h+1
pk
∫ ak
G
(1
G− 1
t
)dt
is clearly nonnegative. On the other hand, it is equal ton∑
k=1
pk
[log t− t
G
]Gak
=n∑
k=1
pk
[logG− log ak − 1 +
akG
]Gak
=
(n∑
k=1
pk
)logG−
n∑k=1
log apkk −n∑
k=1
pk +n∑
k=1
pkakG
=A
G− 1.
From this, A ≥ G. Equality holds if and only if each of these intervalshas length zero, i.e., a1 = a2 = · · · = an(= G).
Exercise
1. Prove that (n+ 1)n ≥ 2n · n! for n = 1, 2, 3, . . . .
2. Prove that for any positive integer n,
1 ≥ nn
(n!)2≥ (4n)n
(n+ 1)2n.
3. Prove that for x > 0,
(1 + x)n+1 ≥((n+ 1)n+1
nn
)x.
4. (Israel Math. Olympiad, 1995.) Prove the inequality
1
kn+
1
kn+ 1+
1
kn+ 2+· · ·+ 1
(k + 1)n− 1≥ n
(n
√k + 1
k− 1
).
![Page 8: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/8.jpg)
424 Inequalities
This clearly suggests using the AG mean inequality. There are n terms on the left. To deal with the−1 on the right, we transfer it to the left. The inequality becomes
1
n
n−1∑j=0
(1 +
1
kn+ j
)≥ n
√k + 1
k.
Note that 1 + 1kn+j
=kn+(j+1)
kn+j. Now it is clear that the AG mean inequality does the job.
The AG means inequality can be rationalized by replacing each xk byxnk . Since G(xn
1 , xn2 . . . , x
nn) = x1x2 · · · xn,
Theorem 27.3. For nonnegative numbers x1, x2, . . . , xn,
Fn(x1, x2, . . . , xn) := xn1 + xn
2 + · · ·+ xnn − n · x1x2 · · · xn ≥ 0.
Equality holds if and only if x1 = x2 = · · · = xn.
Proof. • n = 2: F2(x1, x2) = (x1 − x2)2 ≥ 0.
• n = 4:
F4(x1, x2, x3, x4)
=x41 + x4
2 + x43 + x4
4 − 4x1x2x3x4
=(x41 − 2x2
1x22 + x4
2) + (x43 − 2x2
3x24 + x4
4) + 2(x21x
22 − 2x1x2x3x4 + x2
3x24)
=(x21 − x2
2)2 + (x2
3 − x24)
2 + 2(x1x2 − x3x4)2
≥0.
• n = 3: Let A = A(x1, x2, x3) and G = G(x1, x2, x3). Apply theAG means inequality to the 4 numbers x1, x2, x3 and G. Note that
– A(x1, x2, x3, G) = 3A+G4
,
– G(x1, x2, x3, G) = 4√x1x2x3G =
4√G3 ·G = G.
It follows from 14(3A+G) ≥ G that A ≥ G.
In each of the above, it is clear that equality holds if and only if thex’s are equal.
27.4 Cauchy-Schwarz inequality
Theorem 27.4 (Cauchy-Schwarz inequality). Let x and y be vectors inRn.
|x · y| ≤ ||x|| · ||y||.
![Page 9: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/9.jpg)
27.4 Cauchy-Schwarz inequality 425
Equality holds if and only if x and y are linearly dependent. More ex-plicitly, if x = (x1, x2, . . . , xn) and y = (y1, y2, . . . , yn), then
(x1y1 + x2y2 + · · · xnyn)2 ≤ (x2
1 + x22 + · · ·+ x2
n)(y21 + y22 + · · ·+ y2n).
Proof. For a real number t, consider the vector x− ty, which has squarelength
||x||2 − 2(x · y)t+ ||y||2t2.Since this quadratic in t is always nonnegative, its discriminant |x ·y|2−||x||2||y||2 ≤ 0. Equality holds if and only if there is a real number t forwhich the square length of the vector x − t · y is zero. This is the caseprecisely when x and y are linearly dependent.
An immediate corollary is the triangle inequality in Rn:
Corollary 27.5 (Minkowski inequality). ||x+y|| ≤ ||x||+ ||y||. Equal-ity holds if and only if x and y are linearly dependent.
Proof. ||x+y||2 = ||x||2+2(x·y)+||y||2 ≤ ||x||2+2||x||·||y||+||y||2 =(||x||+ ||y||)2.
Examples
1. For positive numbers x1, x2, . . . , xn,
(x1 + x2 + · · ·+ xn)
(1
x1
+1
x2
+ · · ·+ 1
xn
)≥ n2.
2. Show that 7
1 +1
2+
1
3+ · · ·+ 1
n≥ 2n
n+ 1.
3. The inequality
x21
y1+
x22
y2+ · · ·+ x2
n
yn≥ (x1 + x2 + · · ·+ xn)
2
y1 + y2 + · · ·+ yn
for positive numbers xk and yk, k = 1, 2, . . . , n, 8 follows by ap-plying the Cauchy-Schwarz inequality to the vectors(
x1√y1,x2√y2, . . . ,
xn√yn
)and (
√y1,
√y2, . . . ,
√yn).
7Easy; but the result is too crude, since the right hand side is below 2.8Actually, only the y’s need to be positive.
![Page 10: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/10.jpg)
426 Inequalities
4. (Crux Math., Problem 2113.) Prove the inequality(n∑
k=1
xk
)(n∑
k=1
yk
)≥(
n∑k=1
(xk + yk)
)(n∑
k=1
xkykxk + yk
)
for positive numbers x1, x2, . . . , xn, and y1, y2, . . . , yn.
5. (Crux Math., Problem 1982.)
Determine all sequences a1 ≤ a2 ≤ · · · ≤ an of positive realnumbers such that
n∑k=1
ak = 96,n∑
k=1
a2k = 144,n∑
k=1
a3k = 216.
![Page 11: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/11.jpg)
27.5 Convexity 427
6. Locate the point P inside a given triangle ABC for which the sumof the squares of the distances to the three sides is smallest possible.Suppose the given triangle ABC has side lengths BC = a, CA = b, AB = c, and area �. Letthe distances from P to BC, CA, AB be x, y, z respectively. These satisfy ax+ by + cz = 2�.By the Cauchy-Schwarz inequality,
(x2 + y2 + z2) ≥ (ax+ by + cz)2
a2 + b2 + c2=
4�2
a2 + b2 + c2.
Equality holds if and only if x : y : z = a : b : c. The sum of squares of the distances of a pointto the three sides is smallest when these distances are in the ratio of the corresponding side lengths.Since the actual distances can be determined, there is only one such point P . Here is a constructionof P . Construct squares externally on the sides of the triangle. The outer sides of the square bound alarger triangle A′B′C′. Construct the lines AA′ and CC′ to intersect at a point P . This is the pointthat satisfies the requirement.
PX : XX′ = PC : CC′ = PY : Y Y ′ = PA : AA′ = PZ : ZZ′ =⇒ x : y : z = a : b : c.
The point P also lies on the line BB′. It is called the symmedian point of triangle ABC.
A′C′
B′
P
Y
Y ′
XX′Z
Z′
C
B
A
27.5 Convexity
A real valued function f on an interval I is convex if
f(tx1 + (1− t)x2) ≤ tf(x1) + (1− t)f(x2)
for x1, x2 ∈ I and t ∈ [0, 1].
![Page 12: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/12.jpg)
428 Inequalities
Lemma 27.6. If f is a convex function on an interval I , and x1, x2, . . . ,xn ∈ I , λ1, λ2, . . . , λn ∈ [0, 1] such that
∑nk=1 λk = 1, then
f
(n∑
k=1
λkxk
)≤
n∑k=1
λkf(xk).
Theorem 27.7. Let r > 0. For x,y ∈ Rn, x nonnegative and y positive,n∑
k=1
xr+1k
yrk≥ (
∑nk=1 xk)
r+1
(∑n
k=1 yk)r .
Equality holds if and only if x and y are linearly dependent.
Proof. Apply Lemma 27.6 to the convex function xr+1 on the interval(0,∞). Let zk =
xk
ykfor each k. Set λk =
ykY
, where Y =∑n
k=1 yk. Notethat
∑nk=1 λk = 1. Therefore,(
n∑k=1
xk
Y
)r+1
≤n∑
k=1
ykY
(xk
yk
)r+1
.
Rearranging this, we obtain
Xr+1
Y r≤
n∑k=1
xr+1k
yrk.
1. Let a, b be positive numbers, and x and y be nonnegative numberssatisfying x+ y = 1. Find the least possible value of a
xn + byn
.
Write a = αn+1 and b = βn+1. Since
a
xn+
b
yn=
αn+1
xn+
βn+1
yn≥ (α+ β)n+1
(x+ y)n= (α+ β)n+1 = ( n+1
√a+
n+1√b)n+1,
this latter is the least possible value; it occurs when x : y = α : β = n+1√a :
n+1√b, i.e.,
x =n+1√a
n+1√a+n+1√
b, and y =
n+1√b
n+1√a+n+1√
b.
2. Given positive numbers a1, . . . , an, find the least possible value ofa1xk1
+a2xk2
+ · · ·+ anxkn
subject to x21 + x2
2 + · · ·+ x2n = 1.
3. Prove that for real numbers x > y > z,
2x(y − z) + 2y(z − x) + 2z(x− y) > 0.
![Page 13: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/13.jpg)
27.6 The power mean inequality 429
27.6 The power mean inequality
Throughout this chapter, x denotes a vector (x1, x2, . . . , xn) of nonneg-ative numbers. Let s be a nonzero rational number. We define the s-power mean 9
As(x) =
(∑nk=1 x
sk
n
) 1s
.
Theorem 27.8 (Power mean inequality). For a given x, the s-powermean As(x) is an increasing function of s, i.e., for r < s,(∑n
k=1 xrk
n
) 1r
≤(∑n
k=1 xsk
n
) 1s
.
Equality holds if and only if x1 = x2 = · · · = xn.
Theorem 27.9. 1. lims ��−∞ As(x) = min(x1, x2, . . . , xn);
2. lims ��∞ As(x) = max(x1, x2, . . . , xn);
3. lims �� 0 As(x) = G(x).
Examples
1. For a positive integer k > 1,
1k + 3k + · · ·+ (2n− 1)k > nk+1.
L.S. = n ·Ak(1, 3, . . . , 2n− 1)k > n ·A1(1, 3, . . . , 2n− 1)k = n · nk = nk+1.
2. For positive numbers x, y, z,
x√y + z
+y√z + x
+z√x+ y
>
√x+
√y +
√z√
3.
x√y + z
+y√z + x
+z√x+ y
>x+ y + z√x+ y + z
=√x+ y + z
=√
3A1(x, y, z) ≥√
3A 12(x, y, z) =
√3 ·(√
x+√y +
√z
3
)2
=
√x+
√y +
√z√
3.
9For s < 0, we require x1, x2, . . . , xn to be positive.
![Page 14: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/14.jpg)
430 Inequalities
Remark. Crux Math., Problem 1366 improves the inequality by re-placing
√3 in the denominator by
√2.
3. 3(x2y + y2z + z2x)(xy2 + yz2 + zx2) ≥ xyz(x+ y + z)3.
(x2y + y2z + z2x)(xy2 + yz2 + zx2) ≥ (x2√yz + y2√zx+ z2
√xy)2
=xyz(x
32 + y
32 + z
32
)2
≥ xyz · (3(A3/2)32 )2 ≥ 9xyz · A3
1 =1
3xyz(x+ y + z)3.
4.
5.
27.7 Bernoulli inequality
Theorem 27.10 (Bernoulli inequality). Let n ≥ 2 be a natural number.
(1 + x)n ≥ 1 + nx for x > −1.
1. If n is odd, the Bernoulli inequality is valid for x ≥ −2.
2. If n is even, the Bernoulli inequality is valid for all x ∈ R.
Exercise
1. Prove that for x > 0,
(1 + x)n+1 ≥((n+ 1)n+1
nn
)x.
Crux 2260: This is indeed valid for all n. This inequality is an improvement of the Bernoulli in-
equality since (n+1)n+1
nn =(1 + 1
n
)n+1 · n > e · n.
2.
3.
27.8 Holland’s Inequality
Let x1, x2, . . . , xn be a sequence of nonnegative numbers. For k =1, 2, . . . , n, let
• Ak := A(x1, . . . , xk),
![Page 15: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/15.jpg)
27.8 Holland’s Inequality 431
• Gk := G(x1, . . . , xk).
Theorem 27.11 (Holland’s inequality). A(G1, G2, . . . , Gn) ≥ G(A1, A2, . . . , An).Equality holds if and only if x1 = x2 = · · · = xn.
Lemma 27.12 (Holder inequality). Let ak, bk, k = 1, 2, . . . , n be posi-tive, and p, q > 1 satisfy 1
p+ 1
q= 1.
n∑k=1
akbk ≤(
n∑k=1
apk
) 1p(
n∑k=1
bqk
) 1q
.
Lemma 27.13. For positive numbers a, b, p, q with 0 < p, q < 1, p+q =1,
apbq ≤ pa+ qb.
Proof of Theorem
The inequality is clearly true for n ≤ 2. Let n > 2. For k = 2, . . . , n−1,rewrite
Gk = x1nk
(G
n−kn
k ·Gk−1n
k−1
).
We have(G1 + · · ·+Gn
n
)n
=
⎛⎝x
1n1 ·G
n−1n
1 + · · ·+ x1nk ·G
n−kn
k Gk−1n
k−1 + · · ·+ x1nn ·G
n−1n
n−1
n
⎞⎠n
≤(x1 + · · · xn
n
)⎛⎝G1 +∑n−1
k=2 Gn−kn−1
k Gk−1n−1
k−1 +Gn−1
n
⎞⎠n−1
by Holder’s inequality. Applying Lemma 27.13 to the second factor,with p = n−k
n−1, q = k−1
n−1, we continue
≤(x1 + · · · xn
n
)(G1 +
∑n−1k=2
(n−kn−1
Gk +k−1n−1
Gk−1
)+Gn−1
n
)n−1
=
(x1 + · · · xn
n
)(G1 +G2 + . . . Gn−1
n− 1
)n−1
.
The result now follows by induction.
![Page 16: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/16.jpg)
432 Inequalities
27.9 Miscellaneous inequalities
1. For real numbers a and b,
|a+ b|1 + |a+ b| ≤
|a|1 + |a| +
|b|1 + |b| .
Analysis: Note that each of the terms is of the form 1 − 1x
. This observation reduces to inequalityinto
1− 1
1 + |a| −1
1 + |b| +1
1 + |a+ b| ≥ 0.
The left hand side can be factored if instead of 11+|a+b| , we have 1
(1+|a|)(1+|b|) . To maintain the
proper ordering, we have to make sure that 1 + |a+ b| ≤ (1 + |a|)(1 + |b|). This is clear since
1 + |a+ b| ≤ 1 + |a|+ |b| ≤ 1 + |a|+ |b|+ |a||b| = (1 + |a|)(1 + |b|).We reorganize this into a direct proof:
|a+ b|1 + |a+ b| ≤1− 1
1 + |a+ b|
=2−(1 +
1
1 + |a+ b|)
≤2−(1 +
1
1 + |a|+ |b|+ |ab|)
≤2−(1 +
1
(1 + |a|)(1 + |b|))
=2−(
1
1 + |a| +1
1 + |b|)
=
(1− 1
1 + |a|)
+
(1− 1
1 + |b|)
=|a|
1 + |a| +|b|
1 + |b| .
2. (Crux Math., Problem 1976.) If a, b, c are positive numbers, provethat
a(3a− b
c(a+ b)+
b(3b− c)
a(b+ c)+
c(3c− a)
b(c+ a)≤ a3 + b3 + c3
abc.
3. (Crux Math., Problem 1940.) Show that if x, y, z > 0,
(xy + yz + zx)
(1
(x+ y)2+
1
(y + z)2+
1
(z + x)2
)≥ 9
4.
4.
![Page 17: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/17.jpg)
27.10 A useful variant of AG2 433
27.10 A useful variant of AG2
The AG mean inequality for two positive numbers x and y can be rewrit-ten as
x2
y≥ 2x− y. (AG′
2)
We present some interesting applications of this variant.
1. Let a1, a2, . . . , an be positive numbers. For each k = 1, 2, . . . , n,applying (AG′
2) to ak and ak+1 (stipulating that an+1 = a1), andadding up the n inequalities, we have
n∑k=1
a2kak+1
≥∑
2ak − ak+1 = 2n∑
k=1
ak −n∑
k=1
ak+1 =n∑
k=1
ak.
Corollary: For three positive numbers a b, c, a2
b+ b2
c+ c2
a≥ a+b+c.
2. Consider n positive numbers a1, a2, . . . , an, with sum S. For eachk = 1, 2, . . . , n, applying (AG′
2) to ak and the arithmetic mean ofthe remaining terms, namely, S−ak
n−1, we have
a2k1
n−1(S − ak)
≥ 2ak − S − akn− 1
= · · · = (2n− 1)ak − S
n− 1.
Adding these n inequalites, we obtain
(n−1)n∑
k=1
a2kS − ak
≥n∑
k=1
(2n− 1)ak − S
n− 1=
(2n− 1)S − nS
n− 1= S.
It follows thatn∑
k=1
a2kak+1 + · · ·+ ak−1
≥ a1 + a2 + · · ·+ ann− 1
.
Equality holds if and only if ak = S−akn−1
, i.e., ak = Sn
for each k, .This is equivalent to a1 = a2 = · · · = an.
For three numbers a, b, c, we have
a2
b+ c+
b2
c+ a+
c2
a+ b≥ a+ b+ c
2.
![Page 18: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/18.jpg)
434 Inequalities
3. Let a1, a2, . . . , an be positive numbers such that aa+a2+· · ·+an =1. Show that ∑
k=1
a2k1− ak
≥ 1
n− 1.
4. Let x1, x2, . . . , xn and y1, y2, . . . , yn be two sequences of positivenumbers.
x21
y1+
x22
y2+ · · ·+ x2
n
yn≥ (x1 + x2 + · · ·+ xn)
2
y1 + y2 + · · ·+ yn.
Proof. Let X := x1 + x2 + · · · + xn and Y := y1 + y2 + · · · +yn. Apply (AG′
2) to(YX
)2xk and yk for each k and add up the
inequalities.
5. Let a1, a2, . . . , an be (pairwise) distinct positive integers. Applying(AG′
2) to ak and k for each k, and adding up the inequalities, weobtain
n∑k=1
a2kk
≥n∑
k=1
2ak − k = 2n∑
k=1
ak − Tn.
Since a1, a2, . . . , ak are distinct positive integers, their sum exceedsthe nth triangular number Tk. It follows that
n∑k=1
a2kk
≥ n(n+ 1)
2.
6. (International Mathematical Olympiad, 1978.) Let a1, a2, . . . , anbe (pairwise) distinct positive integers. Show that
n∑k=1
akk2
≥n∑
k=1
1
k.
Exercise
1. (Should this be here?) If a and b are positive numbers such thata+ b = 1, prove that
(a+1
a)2 + (b+
1
b)2 ≥ 25
2.
![Page 19: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/19.jpg)
Chapter 28
Calculus problems
28.1 Philo’s line
Given a point P (a, b) in the first quadrant, find the length of the shortestsegment XY through P , with X , Y on the x- and y-axes respectively.
b
a
Y
X
P
O
θ
In terms of the angle θ between XY and the x-axis, the length of XYis L = a
cos θ+ b
sin θ.
dL
dθ=
a sin θ
cos2 θ− b cos θ
sin2 θ=
a sin3 θ − b cos3 θ
sin2 θ cos2 θ.
dLdθ
= 0 if and only if a sin3 θ = b cos3 θ. Equivalently,
tan3 θ =b
a.
The position of this shortest segment, called Philo’s line, cannot beconstructed with ruler and compass.
![Page 20: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/20.jpg)
436 Calculus problems
28.1.1 Triangle with minimum perimeter
The perimeter of triangle XOY is
P = a+ b+ a tan θ +a
cos θ+ b cot θ +
b
sin θ
= a+ b+a(1 + sin θ)
cos θ+
b(1 + cos θ)
sin θ
dP
dθ=
a(cos2 θ + (1 + sin θ) sin θ)
cos2 θ− b(sin2 θ + (1 + cos θ) cos θ)
sin2 θ
=a(1 + sin θ)
cos2 θ− b(1 + cos θ)
sin2 θ
=a
1− sin θ− b
1− cos θ
=a− b− a cos θ + b sin θ
(1− sin θ)(1− cos θ).
dPdθ
= 0 if and only if
a cos θ − b sin θ = a− b.
Since
d2P
dθ2=
d
dθ
(a
1− sin θ− b
1− cos θ
)=
a cos θ
(1− sin θ)2+
b sin θ
(1− cos θ)2> 0
for acute angle θ, the solution of a cos θ − b sin θ = a − b gives theminimum perimeter P .
This equation can be rewritten in the form
cos(θ + α) =a− b√a2 + b2
with
cosα =a√
a2 + b2and sinα =
b√a2 + b2
.
The position of the segment XY can be constructed with ruler andcompass.
![Page 21: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/21.jpg)
28.1 Philo’s line 437
P (a, b)
Q
Z
X
Y
O
Y ′
X′
αa
θ
b
Without loss of generality, assume a ≥ b.(1) Construct a point Q on the segment OP such that PQ = a− b.
(2) Draw the perpendicular to OP at Q to intersect the circle P (O) at Z(so that angle Y ′PZ is acute).(3) Construct the line PZ to intersect the axes at X and Y respectively.
Triangle XOY has the least perimeter among all triangles with XYcontaining P .
The minimum perimeter is 2(√a+
√b)2.
Exercise
1. Given a cos θ − b sin θ = a− b, find a sin θ + b cos θ.
Solution.
(a sin θ + b cos θ)2 + (a cos θ − b sin θ)2 = a2 + b2
=⇒ (a sin θ + b cos θ)2 = (a2 + b2)− (a− b)2 = 2ab.
2. A coffee filter is formed by gluing two radii of a sector cut outfrom a circle. For what angle of the sector does the coffee filterhave largest capacity?
![Page 22: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/22.jpg)
438 Calculus problems
3. A right pyramid has a square base and a given surface area A. Whatis the largest possible volume?
hl
2b
4. Inscribe in a given cone a cylinder whose volume is largest possi-ble.
2r
hr
5. Find the largest cylinder contained in a sphere of radius R.
r
R
h
6. Find the largest right circular cone contained in a sphere of radiusR.
7. A pencil of length � just fits inside a cylindrical tin. What are thedimensions of the tin if
![Page 23: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/23.jpg)
28.1 Philo’s line 439
r
R
h
(a) the volume is maximum?
(b) the area of the curved surface without lids is maximum?
(c) the area of the curved surface with the bottom lid only is max-imum?
(d) the area of of the curved surface with both lids is maximum?
8. Let P be a point on the graph of a cubic polynomial y = ax3 + bx,(a �= 0). The tangent at P intersects the graph again at Q.
(a) Find the area A between the tangent PQ and the graph.
(b) If the tangent at Q intersects the graph again at R, show thatthe area between the tangent QR and the graph is B = 16A.
9. Show that∫ 1
0x4(1−x)4
1+x2 dx = 227− π.
![Page 24: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/24.jpg)
440 Calculus problems
28.2 Maxima and minima without calculus
1. We have 1000 feet of fencing and wish to make with it a rectangularpen, at one side of a long wall. How can we arrange to surroundthe maximum area?
x
1000− 2x
x
![Page 25: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/25.jpg)
28.2 Maxima and minima without calculus 441
2. A coffee filter is formed by gluing two radii of a sector cut outfrom a circle. For what angle of the sector does the coffee filterhave largest capacity?
Solution. Suppose the circle has radius �. If the coffee filter hasbase radius r and height h, r2 + h2 = �2. Write this as 1
2r2 + 1
2r2 +
h2 = �2. The capacity V of the coffee filter is given by
V 2 =π2
9r4h2 =
4π2
9· r
2
2· r
2
2· h2
≤ 4π2
9·( r2
2+ r2
2+ h2
3
)3
=4π2
9·(�2
3
)3
=4π2�6
243.
This means V ≤ 2π9√3�3. This occurs when r2
2= h2 = �2
3, or
r2 = 23�2. The radius of the cone is
√23= 1
3
√6 of that of the
circle. The angle of the sector is therefore 2π · 13
√6 radians. This
is approximately 294 degrees.
![Page 26: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/26.jpg)
442 Calculus problems
3. A Norman window has a fixed perimeter a. Find the largest possi-ble area.
r
h
2r
![Page 27: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/27.jpg)
28.2 Maxima and minima without calculus 443
4. A right pyramid has a square base and a given surface area A. Whatis the largest possible volume?
hl
2b
![Page 28: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/28.jpg)
444 Calculus problems
5. A tray is constructed from a square metal sheet by dimension a ×a by cutting squares of the same size from the four corners andfolding up the sides. What is the largest possible capacity of thetray?
6. The perimeter of a triangle is a constant 2s. What is the largestpossible area of the triangle?
![Page 29: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/29.jpg)
28.2 Maxima and minima without calculus 445
7. The volume of a cylindrical can is given by V = πr2l and thesurface area by A = 2πr(l + r). If the volume is a constant V ,what is the least possible surface area?
8. Inscribe in a given cone a cylinder whose volume is largest possi-ble.
2r
h
r
![Page 30: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/30.jpg)
446 Calculus problems
9. Find the largest cylinder contained in a sphere of radius R.
r
R
h
![Page 31: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/31.jpg)
28.2 Maxima and minima without calculus 447
10. Find the largest right circular cone contained in a sphere of radiusR.
r
R
h
![Page 32: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/32.jpg)
448 Calculus problems
11. Two corridors of widths a and b meet at a right-angled corner. Whatis the length of the longest ladder which may be carried round thecorner? Assume the workman sufficiently unimaginative to keepthe ladder horizontal).
b
v
a u
![Page 33: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/33.jpg)
Chapter 29
29.1 Quartic polynomials and the golden ratio
Suppose the graph of a quartic polynomial f(x) has two points of inflec-tions A and B. If the line AB intersects the graph of f(x) at P and Q(in the order P , Q, B, Q), then A divides BP and B divides AQ in thegolden ratio.
A
B
Q
P
Proof. Without loss of generality we may assume f(x) is monic, withf(0) = 0, andthe x-coordinates of the points of inflections equal to ±a for some a.We write f ′′(x) = 12(x2 − a2).Integrating, we obtain
f(x) = x2(x2 − 6a2) + bx
for some constant b.
![Page 34: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/34.jpg)
450
Since
f(a) = b · a− 5a4,
f(−a) = b(−a)− 5a4,
the line containing the points of inflection is clearly
y = bx− 5a4.
This line intersects the graph of f(x) at four points given by
x2(x2 − 6a2) + bx = bx− 5a4,
orx2(x2 − 6a2) + 5a4 = 0 =⇒ (x2 − a2)(x2 − 5a2) = 0.
Two of these roots are ±a, corresponding to the two points of inflection.The other two roots are ±√
5a. These are the x-coordinates of P and Q.It follows that A divides BP and B divides AQ in the golden ratio.
(1) There are three points on the graph y = f(x) where the tangentsare parallel to the line AB.These are given by 4x3 − 12a2x+ b = b, i.e., x = 0, x = ±√
3a.(i) The tangents at x = ε
√3a, ε = ±1, are the same line y + 9a4 +
ε√3ab = b(x+ ε
√3a), i.e., y = bx− 9a4.
The area of the hump is 8√3
5a5, independent of b.
A
B
Q
P
(ii) The tangent at x = 0 is the line y = bx. It intersects the graphagain at x = ±√
6a.The area of the double hump is 2
√6
5a5. The two parts are equal in
area.
![Page 35: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/35.jpg)
29.1 Quartic polynomials and the golden ratio 451
A
B
Q
P
The area bounded by the graph of y = f(x) and the line AB(i) between A and P is equal that between B and Q,(ii) between A and B is equal to the sum of the areas between A and P ,and between B and Q.
A
B
Q
P
![Page 36: Inequalities - Florida Atlantic Universitymath.fau.edu/yiu/MPS2016/PSRM2016K.pdf · 2016. 3. 30. · Inequalities 27.1 The inequality AGH2 Theorem 27.1. For positive numbers xand](https://reader035.vdocument.in/reader035/viewer/2022062610/611c904e7028534cd450dc16/html5/thumbnails/36.jpg)
452