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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
.
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.
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.
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).
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 ?
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.
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
).
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||.
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.
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.
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].
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.
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.
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),
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.
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.
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.
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.
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.
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.
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?
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
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.
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
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.
442 Calculus problems
3. A Norman window has a fixed perimeter a. Find the largest possi-ble area.
r
h
2r
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
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?
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
446 Calculus problems
9. Find the largest cylinder contained in a sphere of radius R.
r
R
h
28.2 Maxima and minima without calculus 447
10. Find the largest right circular cone contained in a sphere of radiusR.
r
R
h
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
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.
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.
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
452
