Kvant Inequalities

Selected Inequalities from Êâàíò ÆóðíàëÊâàíò Translators - Mathematical Community MathVn

http://mathvn.org

1

Information

• Translated from the mathematical magazine Êâàíò, all issues from 1970 to 2008

• Typeset by LATEX2ε

• Copyright c©2008 Mathematical Community MathVn - http://mathvn.org

• This paper will be updated and completed in the next version

2

PROBLEMS

7. Let a, b, c be the lengths of the sides of a triangle, prove that

a

b + c− a+

b

c + a− b+

c

a + b− c≥ 3

13. Let d be the difference of the maximum and the minimum from n distinct realnumbers a1, a2, ..., an, and s be the sum of n(n − 1)/2 absolutes of differences of all pairsmade from these numbers

∑

i<j

|ai − aj |. Prove that (n− 1)d ≤ s ≤ n2d/4

23. For every natural number n > 1, prove that the quotient between the product offirst n even numbers with the product of first n odd numbers is greater than

√8n/3 and

less than√

4n.

33. For every natural number n > 1000, prove that the sum of all remainders fromdivision of 2n for 1, 2, .., n greater than 2n.

51. If product of three positive numbers is 3 and their sum greater than the sum of theirinverses. Prove that exist at least a number from these number greater than 1.

90. For x1 < x2 < ... < xn be natural numbers, prove that√

x2 − x1

x2+√

x3 − x2

x3+ ... +

√xn − xn−1

xn< 1 +

12

+13

+ ... +1n2

105 Let S(n) be the sum of all digits of a natural number n. Prove that S(8n)S(n)

≥ 18.

For which natural number k in order that exists positive number ck satisfied the inequalityS(kn)S(n)

≥ ck for all natural number n? Find the greatest value of ck.

121. Let a1, a2, ..., an be real numbers. Prove that exist a natural number k ≤ n suchthat all the values of ak,

ak + ak−1

2, ak + ak−1 + ak−2

3,...,ak + ... + a2 + a1

kare not greater

than arithmetic mean of a1, a2, ..., an.

157. The sum of n positive numbers x1, x2, ..., xn is 1. S is the greatest number fromx1

1 + x1, x2

1 + x1 + x2,..., xn

1 + x1 + ... + xn. Find the least value of S.

162. Let a1 < a2 < ... < an be an increasing sequence of natural numbers such thatfor every natural number m then m ∈ (an) or have the unique form m = ak + al, wherek, l ∈ N. Prove that an ≤ n2 for all n ∈ N

182. Let a1, a2, ..., an be potisive numbers, prove thata1

a2 + a3 + ... + an+

a2

a1 + a3 + ... + an+ ... +

an

a1 + a2 + .... + an−1≥ n

n− 1

3

208. The difference between the greatest and the least value of numbers x1, x2, ..., xn

equals 1. Find the greatest and the least value of x1 + x2 + ... + xk

k, where k = 1, 2, ..., n

218. Let x1, x2, x3, x4, x5 be positive numbers, prove that

(x1 + x2 + x3 + x4 + x5)2 ≥ 4(x1x2 + x2x3 + x3x4 + x4x5 + x5x1)

224. Let a1, a2, ..., an and b1, b2, ..., bn be real numbers. Prove that when one from fol-lowing conditions occur:

i. If ai < aj then bi ≤ bj

ii. If ai ≤ a1 + a2 + ... + an

n< aj then bi ≤ bj

for all numbers i and j, then

(a1 + a2 + ... + an)(b1 + b2 + ... + bn) ≤ n(a1b1 + a2b2 + ... + anbn)

257. Which value of natural number n ≥ 2 then

x21 + x2

2 + ... + x2n ≥ p(x1x2 + x2x3 + ... + xn−1xn)

where

a. p = 1

b. p =43

c. p =65

Due to Nguyen Cuong Qui, Vietnam - 1974

294. Let a, b, c, d, x, y, u, v be real numbers such that abcd > 0. Prove that

(ax+bu)(av+by)(cx+dv)(cu+dy) ≥ (àñuvx+bcuxy+advxy+bduvy)(acx+bcu+adv+bdy)

308. Let a1cosx + a2cos2x + ... + ancosnx ≥ −1 for every x. Prove that

a1 + a2 + ... + an ≤ n

360. Let a1, a2, ..., an be real numbers such that |a1| = 1 and |ak+1| = |ak + 1| fork = 1, 2, ..., n− 1. Find the least value of |a1 + a2 + ... + an| if

a. n = 1975.

b. n = 1976.

374. Let a, b, c be positive numbers such that a > c, b > c. Prove that√

c(a− c) +√

c(b− c) ≤√

ab

4

418. Let n be natural number, prove that

n( n√

n + 1− 1) ≤ 1 +12

+13

+ ... +1n≤ 1 + n(1− 1

n√

n).

423. Let x, y, z be real numbers, prove that

(x2 + y2 − z2)(x2 + z2 − y2)(y2 + z2 − x2) ≤ (x + y − z)2(x + z − y)2(y + z − x)2

469. Prove that if the equation xn + a1xn−1 + ... + ak−1x

n−k+1 + ak+1xn−k−1 + ... +

an−1x + an = 0 has n distinct roots in R then ak+1ak−1 < 0.

503. Let a0, a1, a2, ..., a2n be real numbers such that ak ≥ ak−1 + ak+1

2, k = 1, 2, ..., 2n−

1. Prove thata1 + a3 + ... + a2n−1

n≥ a0 + a2 + ... + a2n

n + 1when does the equation hold?

506. Let a, b, c, d be positive numbers, prove that

a4 + b4 + c4 + d4 + 2abcd ≥ a2b2 + a2c2 + a2d2 + b2c2 + b2d2 + c2d2.

507. Let a1 < a2 < ... < an < 2n be natural numbers and n > 5. Prove that

a. min1≤i<j≤n

lcm[ai, aj ] ≤ 6([n

2] + 1).

b. max1≤i<j≤n

gcd(ai, aj) >38n

147− c, where c does not depend on n.

More than that, we can not replace 6 (at 507.a.) by a smaller number and can not replace38147

by a greater number (at 507.b.).

510. Assume that 0 < α < π and n is a natural number. Prove that

sinα +12

sin 2α +13

sin 3α + ... +1n

sinnα > 0

518. Let x1, x2, ..., xn be positive numbers in [a, b], 0 < a < b. Prove that

(x1 + x2 + ... + xn)(1x1

+1x2

+ ... +1xn

) ≤ (a + b)2n2

4ab

527. Assume that x1, x2, ..., xn ∈ [0, 1] and n ≥ 3. Prove that

x1 + x2 + ... + xn − x1x2 − ...− xn−1xn − xnx1 ≤[n

2

]

532. Assume that an =√

n− 1 +√

n and bn =√

4n + 2. Prove that

a. [an] = [bn]

b. 0 < bn − an < 116n

√n

for n = 1, 2, 3...

5

543. Given ρ(x, y) =|x− y|√

(1 + x2)(1 + y2). Prove that for every real numbers a, b, c then

ρ(a, c) ≤ ρ(a, b) + ρ(b, c)

559. Let a, b, c be the lengths of the sides of a triangle, prove that∣∣∣∣a

b+

b

c+

c

a− b

a− c

b− a

c

∣∣∣∣ < 1

565. Let a1, a2, an be positive numbers, bk is arithmetic mean of all possible products ofk numbers from a1, a2, ..., an. Example,

b1 =a1 + a2 + ... + an

n,

b2 =a1a2 + a1a3 + ... + an−1an

n(n− 1)/2, ...,

bn = a1a2...an.

Prove that

a. b1 ≥√

b2

b. b2k ≥ bk−1bk+1, k = 2, 3, ..., n− 1

c. k√

bk ≥ k+1√

bk+1, k = 2, 3, ..., n− 1.

571. Let x1, x2, ..., xn, ... be a decreasing sequence of positive numbers such thatx1

1+

x4

2+ ... +

xn2

n≤ 1

For all natural number n, prove thatx1

1+

x2

2+ ... +

xn

n< 2

and can not replace 2 by a smaller number.

579. Let x1, x2, ..., xn be real numbers in [0; 1], prove that(x1 + x2 + ... + xn + 1)2 ≥ 4(x2

1 + x22 + ... + x2

n)

590. Let x be real number and n be natural number, prove that

| cosx|+ | cos 2x|+ | cos 4x|+ ... + | cos 2nx| ≥ n

2

597. Let xn = 1 +12

+13

+ ... +1n

. Prove that exists γ = limm→∞(xn − lnn) and

γ < xm + xn − xmn ≤ 1 for every natural numbers m,n.

620. Let x1, x2, ..., xn be real numbers such that x21 + x2

2 + ... + x2n = 1. Prove that the

sum of all 2n absolute values of terms, which have form ±x1 ± x2 ± ... ± xn (all possiblecombinations of signs + and −) is not greater than 2n.

6

647. Let 12 ≤ a ≤ b, prove that

(b2 − a2

2

)2

≥√

a2 + b2

2− a + b

2.

666. Prove that the least common multiple of n natural numbers a1 < a2 < ... < an isnot less than na1.

718. Find the greatest value of m2 + n2, where m,n are natural numbers such thatm ≤ 1981, n ≤ 1981 and n2 −mn−m2 = ±1.

727. Prove that a2 + b2 + c2 + 2abc < 2, where a, b, c are the lengths of the sides of atriangle which has perimeter is 2.

749. a. Let x1, x2, x3 be positive numbers, prove that

x1

x2 + x3+

x2

x3 + x1+

x3

x1 + x2≥ 3

2

b. Let x1, x2, ..., xn be positive numbers, n ≤ 4. Prove thatx1

xn + x2+

x2

x1 + x3+ ... +

xn−1

xn−2 + xn+

xn

xn−1 + x1+

x3

x1 + x2≥ 2

The equality only happens with n = 4.

c. Prove that with n > 4 then at the inequality at point b. can not replace number 2 bya larger other.

762. Let a, b, c be positive numbers, prove that

a + b + c ≤ a2 + b2

2c+

b2 + c2

2a+

c2 + a2

2b≤ a3

bc+

b3

ca+

c3

ab

795. Let σ(n) be the sum of all divisors of the natural number n. Prove that existsinfinitely natural numbers n satisfying the condition:

a. σ(n) > 2n

b. σ(n) > 3n

For every natural number n, prove the inequalities

a. σ(n) < log(2n + 1)

b. σ(n) < ln(n + 1).

812. For every natural number n, prove that12

+1

3√

2+

14√

3+ ... +

1(n + 1)

√n

< 2

7

835 Let a, b, c be the lengths of the sides of a triangle, prove that

a− b

a + b+

b− c

b + c+

c− a

c + a<

18

840. a. Let a, b, c be the lengths of the sides of a triangle, prove that

a2b(a− b) + b2c(b− c) + c2a(c− a) ≥ 0

b. Let a, b, c be positive numbers. Prove that

a3c + b3c + c3a ≥ a2bc + b2ca + c2ab

865. Let a0 < a1 < ... < an−1 < an be natural numbers. Prove that1

[a0, a1]+

1[a1, a2]

+ ... +1

[an−1, an]≤ 1− 1

2n

[a, b] is the least common multiple of natural number a, b.

915. Let a, b, c, d be positive number, prove that

a

b + c+

b

c + d+

c

d + a+

d

a + b≥ 2

919. Prove that 9 <

∫ 3

0

4√

x4 + 1dx +∫ 3

1

4√

x4 − 1dx < 9, 0001.

969. Let a, b, c be positive numbers. Prove that

a3

a2 + ab + b2+

b3

b2 + bc + c2+

c3

c2 + ca + a2≥ a + b + c

3.

986. Let a, b be positive numbers, prove that 2√

a + 3 3√

b ≥ 5 5√

ab.

994.Find the greatest value of k, so that the equality

a4 + b4 + c4 + abc(a + b + c) ≥ k(ab + bc + ca)2

is always true for all numbers a, b, c

999.a. Let a1, a2, ..., an be positive number, prove that the inequality

1a1

+2

a1 + a2+ ... +

n

a1 + a2 + ... + an< 4

(1a1

+1a2

+ ... +1an

)

a. Prove that, can replace number 4 at point a. by number 2

b. Prove that, can not replace number 2 by a smaller other.

1011. For n positive numbers a1 ≥ a2 ≥ ... ≥ an, prove that

a. a21 − a2

2 + a23 ≥ (a1 − a2 + a3)2.

8

b. a21 − a2

2 + a23 − a2

4 ≥ (a1 − a2 + a3 − a4)2.

c. a21−a2

2 + ...+(−1)n−2a2n−1 +(−1)n−1a2

n ≥ (a1−a2 + ...+(−1)n−2an−1 +(−1)n−1an)2

1150. Let a1, a2, ..., an be positive numbers. Prove that

(a1 + a2 + ... + an)2

2(a21 + a2

2 + ... + a2n)≤ a1

a2 + a3+

a2

a3 + a4+ ... +

an−1

an + a1+

an

a1 + a2

1171. Assume that hn = 1 +12

+ ... +1n, prove the inequality

1h2

1

+1

2h22

+1

3h23

+ ... +1

nh2n

< 2

1177. Let x1, x2, ..., xn be positive numbers and not greater than 1. Prove that

(1 + x1)1/x2(1 + x2)1/x3(1 + x3)1/x4 ...(1 + xn)1/x1 ≥ 2n

1193. Prove that

ax + by + cz +√

(a2 + b2 + c2)(x2 + y2 + z2) ≥ 23(a + b + c)(x + y + z)

1207. Let x, y be real numbers and m be natural number. Prove that

(x2 + y2)m ≥ 2mxmym + (xm − ym)2

1219. Let x1, x2, ..., xn be positive numbers, n > 1. Prove that

(s− x1)x1 + (s− x2)x2 + ... + (s− xn)xn > n− 1

where s = x1 + x2 + ... + xn.1228. Let a, b, c be positive numbers which be not greater than 1. Prove that

a

bc + 1+

b

ca + 1+

c

ab + 1≤ 2

1272. Let a1, a2, ..., an be positive numbers and their sum is 1. Prove that(

1a2

1

− 1)(

1a2

2

− 1)

...

(1a2

n

− 1)≥ (n2 − 1)n

1277. For n positive numbers a1, a2, ..., an, prove that√

a1 + a2

a3+

√a2 + a3

a4+ ... +

√an−1 + an

a1+

√an + a1

a2≥ n

√2

1323. Let x, y be positive numbers, prove that x.2y + y.2x ≥ x + y.

1328. For x1, x2, ..., xn ∈ [−1, 1] and x31 + x3

2 + ... + x3n = 0, prove that

x1 + x2 + ... + xn ≤ n

3

9

1336. Let m,n be natural number, prove that

1n√

m + 1+

1m√

n + 1> 1

1364. Let a, b, c be non-negative numbers and a + b + c = 1. Prove that

a. 4a3 + 4b3 + 4d3 + 15abc ≥ 1.

b. a3 + b3 + c3 + abc ≥ min{14,19

+d

27}.

1394. Let a1, a2, ..., an, b1, ..., bn be positive numbers. Prove that the sum of all numberwhich have form akbk

ak + bk, 1 ≤ k ≤ n is not greater than AB

A + B, where A = a1 +a2 + ...+an

and B = b1 + b2 + ... + bn.

1402. For 0 < x1 ≤ x2 ≤ ... ≤ xn and n > 2, prove thatx1

x2+

x2

x3+ ... +

xn

x1≥ x2

x1+

x3

x2+ ... +

x1

xn

1404. Let x, y, z be real numbers which satisfy x + y + z = 0 and xyz = 2. Find thegreatest value of x2

y+

y2

z+

z2

x.

1451. Let a, b be natural numbers in order that a + 1b

+b + 1

ais also natural number.

Prove that(lcm[a, b])2 ≤ a + b

1467. For 0 < a1 < a2 < ... < am ≤ n be natural numbers in order that the sum of allnumber which have form ar + as, 1 ≤ r ≤ s ≤ m is greater than n or its value is in the set{a1, a2, ..., am}. Prove that

a1 + a2 + ... + am

m≥ n + 1

21485. For 0 ≤ x1 ≤ x2 ≤ ... ≤ xn. Prove that the value of

xk2(x1 − x3) + xk

3(x2 − x4) + ... + xk1(xn − x2)

is non-negative if k > 1 and is non-positive if 0 < k < 1.

1492. Assume that s(n) = 11 + 22 + ... + nn, n > 3. Prove thata. 3s(n) > (n + 1)n.

b. 2s(n) < (n + 1)n.

c. 1nn

>1

s(n)+

1s(n + 1)

+1

s(n + 3)....

1526. Let a, b, c be positive numbers and their product is 1. Prove that

1a3(b + c)

+1

b3(c + a)+

1c3(a + b)

≥ 32

10

1597. For a, b > 0 and abc = 1, prove thata. 1

1 + 2a+

11 + 2b

+1

1 + 2c≥ 1.

b. 11 + a + b

+1

1 + b + c+

11 + c + a

≤ 1.

1619. Let x, y, z be real numbers which satisfy x2 + xy + y2 = 3 and y2 + yz + z2 = 16.Find the greatest value of xy + yz + zx.

1627. Let x1, x2, ..., xn be real numbers which satisfy |x1 + x2... + xn| = 1 and |xk| ≤n + 1

2, k = 1, 2, ..., n. Prove that exists a permutation y1, y2, ..., yn of x1, x2, ..., xn which

satisfy the inequality|y1 + 2y2 + ... + nyn| ≤ n + 1


a2 + 2bc

b2 + c2+

b2 + 2ca

c2 + a2+

c2 + 2ab

a2 + b2> 3

1699. For every natural number n, prove that

{√

1}+ {√

2}+ ... + {√

n2} ≤ n2 − 12

.

1701. If x, y > 0 and x2 + y3 ≥ x3 + y4, prove that x3 + y3 ≤ 2.

1710. Let p, q, r, x, y, z be positive numbers, such that p + q + r = 1 and xpyqzr = 1.Prove that

p2x2

qy + rz+

q2y2

px + rz+

r2z2

px + qy≥ 1

2

1804. Let a, b, c be positive numbers, prove thata√

a2 + 8bc+

b√b2 + 8ac

+c√

c2 + 8ab≥ 1

1818. Let a, b, c be positive numbers, prove that√

a

b + c+

√b

a + c+

√c

a + b> 2

1821. For every natural number n, prove that∣∣∣{n

1} − {n

2}+ {n

3} − ... + (−1)n{n

n}∣∣∣ <

√2n

1826. Let a, b, c be positive numbers such that 1a

+1b

+1c≥ a + b + c. Prove that

a + b + c ≥ 3abc.

11

1834. Prove that

a. x6y6 + y6z6 + z6x6 + 3x4y4z4 ≥ 2(x3 + y3 + z3)x3y3z3

b. x6 + y6 + z6 + 3x2y2z2 ≥ 2(x3y3 + y3z3 + z3x3).

1839. For 0 < x < π4 , then

a. (cosx)cox2x > (sinx)sin2x

b. (cosx)cox4x < (sinx)sin4x.

1846. For every natural numbers n, and k ≤ n then

(1 +1n

)k ≤ n2 + nk + k2

n2

1849. Let p be a prime number such that p2 = 2n.3n + 1, with m,n are non-negativeintegers. Prove that p < 18

1869. Let x, y be positive number such that x 6= y and xn +1

xm= yn +

1ym

, where m,n

are natural numbers. Prove that

x2 + y2 >n+m

√169

1881. Let a, b, c be positive numbers and their sum is 1. Prove that1

1− a+

11− b

+1

1− c≥ 2

1 + a+

21 + b

+2

1 + c

1904. Let a, b, c be natural numbers such that a(b2+c2) = 2b2c. Prove that 2b ≤ c+a√

a.

1913. If 0 ≤ x ≤ y ≤ 1 prove that

2√

(1− x2)(1− y2) ≤ 2(1− x)(1− y) + 1.

1917. Let a, p, q be natural numbers such that ap + 1 is a multiple of q, and aq + 1 is amultiple of p. Prove that 2a(p + q) > pq.

1928. Let x1, x2, ..., xn be positive number, such that the difference between the maxi-mum number and the minimum number is not larger than 2. Prove that

x1 + x2 + ... + xn ≤√

1 + x1x2 +√

1 + x2x3 + ... +√

1 + xnx1 ≤ x1 + ... + xn + n.

1938. Let x1, x2, ..., xn be real numbers, prove that

max{x1, x2, ..., xn,−x1 − x2 − ...− xn} ≥ |x1|+ |x2|+ ... + |xn|2n− 1

1986. Let x1 ≤ x2 ≤ ... ≤ xn ≤ y1 ≤ y2 ≤ ... ≤ yn. Prove that

(x1 + x2 + ... + xn + y1 + y2 + ... + yn)2 ≥ 4n(x1y1 + x2y2 + ... + xnyn).

12

2002. Let a, b, c be positive numbers such that a + b + c = 1. Prove that1a

+1b

+1c≥ 25

1 + 48abc

2042. Prove thattansin x +cotcos x ≥ 2, 0 < x <

π

2

13

SOLUTION


a

b + c− a+

b

c + a− b+

c

a + b− c≥ 3

Proof. From the hypothesis we can set a = y + z, b = z + x, c = x + y with x, y, z > 0. Theinequality becomes

y + z

x+

z + x

y+

x + y

z≥ 6

By the AM-GM inequality, we have

y + z

x+

z + x

y+

x + y

z≥ 6 6

√y

x× z

x× z

y× x

y× x

z× y

z= 6

Thus we are done. Equality holds when x = y = z or a = b = c.

∇13. Let d be the difference of the maximum and the minimum from n distinct real numbersa1, a2, ..., an, and s be the sum of n(n− 1)/2 absolutes of differences of all pairs made fromthese numbers

∑i<j |ai − aj |. Prove that

(n− 1)d ≤ s ≤ n2d/4

Proof. Assume that a1 < a2 < ... < an and denote dk = ak+1 − ak, we have

d = an − a1 = d1 + d2 + ... + dn−1

|aj − ai| = di + di+1 + ... + dj−1

s =∑

i<j

|ai − aj | =n−1∑

k=1

k(n− k)dk

Notice that k(n−k) ≥ n−1 and k(n−k) ≤ n2

4for k = 1, 2, ..., n−1. Thus, the inequality

has been proved!

∇23. For every natural number n > 1, prove that the quotient between the product of first neven numbers with the product of first n odd numbers is greater than

√8n/3 and less than√

4n.

Proof. Denote pn = 1..3.5...(2n−1)2.4.6...(2n) . We have

p2n =

12

32

2.452

4.6...

(2n− 1)2

(2n− 2)2n.

12n

For k > 1, notice that (2k−1)2

(2k−2)2k = (2k−1)2

(2k−1)2−1> 1. Thus p2

n ≥ 14n .

14

In another way, we also have

p2n =

322

.3.542

.5.762

...(2n− 3)(2n− 1)

(2n− 2)2.2n− 1(2n)2

Finally, the inequality p2n ≤ 3

8n is a consequence of (2k−1)(2k+1)(2k)2

= (2k)2−1(2k)2

< 1.

∇1692. Let a, b, c be the lengths of the sides of a triangle, prove that

a2 + 2bc

b2 + c2+

b2 + 2ca

c2 + a2+

c2 + 2ab

a2 + b2> 3

Proof. Because a, b, c are the lengths of the sides of a triangle so a+b−c, b+c−a, c+a−b > 0.We have

LHS −RHS =∑cyc

a2 − (b− c)2

b2 + c2=

∑cyc

(a− b + c)(a + b− c)b2 + c2

> 0

∇1710. Let p, q, r, x, y, z be positive numbers, such that p + q + r = 1 and xpyqzr = 1. Provethat

p2x2

qy + rz+

q2y2

px + rz+

r2z2

px + qy≥ 1

2

Proof. By the Cauchy Schwarz inequality, we have

p2x2

qy + rz+

q2y2

px + rz+

r2z2

px + qy≥ (px + qy + rz)2

2(px + qy + zx)=

px + qy + rz

2

By the Weighted AM-GM inequality,

px + qy + rz ≥ xpyqzr = 1

So the inequality is proved. Equality holds if and only if p = q = r =13and x = y = z = 1

∇1804. Let a, b, c be positive numbers, prove that

a√a2 + 8bc

+b√

b2 + 8ac+

c√c2 + 8ab

≥ 1

∇

1818. Let a, b, c be positive numbers, prove that√

a

b + c+

√b

a + c+

√c

a + b> 2

∇

15

1821. For every natural number n, prove that∣∣∣{n

1} − {n

2}+ {n

3} − ... + (−1)n{n

n}∣∣∣ <

√2n

1826. Let a, b, c be positive numbers such that 1a

+1b

+1c≥ a + b + c. Prove that

a + b + c ≥ 3abc.

Proof. We will prove that

(a + b + c)2 ≥ 3abc

(1a

+1b

+1c

)≥ 3abc(a + b + c)

We have3abc

(1a

+1b

+1c

)= 3(ab + bc + ca) ≤ (a + b + c)2

So we are done. Equality holds when a = b = c = 1

∇1834. Prove that

a. x6y6 + y6z6 + z6x6 + 3x4y4z4 ≥ 2(x3 + y3 + z3)x3y3z3

b. x6 + y6 + z6 + 3x2y2z2 ≥ 2(x3y3 + y3z3 + z3x3).

Proof. We will prove part b, part a is equivalent to part b if we set xy = a, yz = b, zx = c.By the Schur inequality we have:

x6 + y6 + z6 + 3x2y2z2 ≥ x2y2(x2 + y2) + y2z2(y2 + z2) + z2x2(z2 + x2)

and

x2y2(x2 +y2)+y2z2(y2 +z2)+z2x2(z2 +x2)−2(x3y3 +y3z3 +z3x3) =∑cyc

x2y2(x− y)2 ≥ 0

End this proof. Equality hold if and only if x = y = z.

∇1839. For 0 < x < π

4 , then

a. (cosx)cos x2x > (sinx)sin2 x

b. (cosx)cos x4x < (sinx)sin4 x.

Proof. We use the function f(y) = cosy x − siny x for y ≥ 0 and 0 < x < π4 . We have

f(0) = 0 and f(y) > 0 for y > 0, limy→∞f(y) = 0. Moreover,

f ′(y) = cosy x ln(cosx)− siny x ln(sinx) = cosy x(ln(cosx)− tany x ln(sinx))

16

Notice that f ′(y) only has a real root for y > 0 since g(y) = tany x is monotone function.We also have f(2) = f(2)(cos2 x + sin2 x) = f(4). Hence f ′(2) > 0 and f ′(4) < 0.

The inequalities f ′(2) > 0 and f ′(4) < 0 have the following equivalent forms:cos2 x ln(cosx) > sin2 x ln(sinx)

cos4 x ln(cosx) < sin4 x ln(sinx)

These complete our proof!∇

1846. For every natural numbers n, and k ≤ n then

(1 +1n

)k ≤ n2 + nk + k2

n2

Proof. We use induction to prove this inequality. With k = 1 then the proposition isobvious. Now assume that the proposition is true for all values which are not greater thank. We have(

1 +1n

)k+1

≤(

1 +k

n+

k2

n2

)(1 +

1n

)= 1+

k + 1n

+k2 + k

n2+

k2

n3< 1+

k + 1n

+(k + 1)2

n2

By the principle of induction we receive the proof.∇

1869. Let x, y be positive number such that x 6= y and xn +1

xm= yn +

1ym

, where m,n

are natural numbers. Prove that

x2 + y2 >n+m

√169

Proof. Assume that x = a sin t and y = a cos t, where a > 0, t ∈ (0, π4 )∪ (π

2 ). The conditionof our problem becomes

an+m sinm t cosm t.sinn t− cosn t

sinm t− cosm t= 1 (∗)

Without loss of generality, suppose that t ∈ (π4 , π

2 ). In this case sink+2 t − cosk+2 t ≤sink− cosk t for k ≥ 2. Therefore, we easily realize that

sinn t− cosn t <32(sin t− cos t) (∗∗)

.Besides of this, sink t cosk t

sink t− cosk t≤ sink−2 t cosk−2 t

sink−2 t− cosk−2 tfor k > 2

and clearly sin2 t cos2 t

sin2 t− cos2 t<

sin t cos t

sin t− cos thence

sinm t cosm t

sinm t− cosm t≤ sin t cos t

sin t− cos t(∗ ∗ ∗)

From (∗∗) and (∗ ∗ ∗), we have

sinm t cosm t.sinn t− cosn t

sinm t− cosm t<

32

sin t cos t <34

Since (∗) and notice that x2 + y2 = a2, then we receive the proof.

17

∇1881. Let a, b, c be positive numbers and their sum is 1. Prove that

11− a

+1

1− b+

11− c

≥ 21 + a

+2

1 + b+

21 + c

Proof. By the Cauchy-Schwarz inequality, we have:1x

+1y≥ 4

x + y∀x, y > 0

So we get1

1− a+

11− b

≥ 42− a− b

=4

1 + cSimilarly,

11− b

+1

1− c≥ 4

1 + a,

11− c

+1

1− a≥ 4

1 + b

Summing up these inequalities, we are done. Equality holds when a = b = c =13

∇1904. Let a, b, c be natural numbers such that a(b2 + c2) = 2b2c. Prove that 2b ≤ c + a

√a.

Proof. Denote t = cb or c = bt. We have a = 2bt

1+t2and c = bt = a(1+t2)

2

Assume that t = mn , with gcd(m,n) = 1. Thus a(m2+n2) = 2bmn and a(m2+n2) = 2cn2.

Notice that gcd(m2 + n2,m) = 1 and gcd(m2 + n2, n) = 1, hence a is a multiple of mn2

or a = kmn2. Thus, we have 2b = kn(m2 + n2) and 2c = km(m2 + n2).

Therefore, by the AM-GM inequality:

2b = km.mn + kn3 ≤ km.m2 + n2

2+ k3/2m3/2n3 = c + a3/2

So we are done!∇

1913. If 0 ≤ x ≤ y ≤ 1 prove that2√

(1− x2)(1− y2) ≤ 2(1− x)(1− y) + 1.

First proof. Because 0 ≤ x ≤ y ≤ 1 so we can set x = sin a, y = sin b with a, b ∈[0,

π

2

].

The inequality becomes2 cos a cos b ≤ 2(1− sin a)(1− sin b) + 1 ⇐⇒ (cos a− cos b)2 + (sin a + sin b− 1)2 ≥ 0

Which is clearly true. Equality holds when x = y =12Second proof. By the AM-GM inequality, we have

2√

(1− x2)(1− y2) ≤ (1− x2) + (1− y2) = 2− x2 − y2

So we need to prove that2− x2 − y2 ≤ 2(1− x)(1− y) + 1 ⇐⇒ (x + y − 1)2 ≥ 0

Q.E.D

18

∇1917. Let a, p, q be natural numbers such that ap + 1 is a multiple of q, and aq + 1 is amultiple of p. Prove that 2a(p + q) > pq.

Proof. Assume that d = gcd(p, q), then d | ap + 1 and d | ap. Hence d | ap + 1− ap = 1 orp and q are relatively prime. Therefore, number ap + aq + 1 which is multiple of p and q, isalso multiple of pq. Thus, a(p + q) ≥ pq − 1. Clearly a(p + q) > 1, so 2a(p + q) > pq.

∇1928. Let x1, x2, ..., xn be positive number, such that the difference between the maximumnumber and the minimum number is not larger than 2. Prove that

x1 + x2 + ... + xn ≤√

1 + x1x2 +√

1 + x2x3 + ... +√

1 + xnx1 ≤ x1 + ... + xn + n.

Proof. From the hypothesis we have |xi − xj | ≤ 2 ∀i, j ∈ {1, · · · , n}. We will prove that

xi + xj ≤ 2√

1 + xixj ≤ xi + xj + 2

which is equivalent to

(xi + xj)2 ≤ 4(1 + xixj) ≤ (xi + xj + 2)2

⇐⇒ (xi − xj)2 ≤ 4 and (xi − xj)

2 + 4(xi + xj) ≥ 0 which is true. Applying the aboveinequality we are done.

∇1938. Let x1, x2, ..., xn be real numbers, prove that

max{x1, x2, ..., xn,−x1 − x2 − ...− xn} ≥ |x1|+ |x2|+ ... + |xn|2n− 1

Proof. If all of numbers x1, x2, ..., xn are non-negative, the inequality is clearly true. With-out loss of generality, assume that x1, x2, ..., xk are negative and xk+1, ..., xn are non-negativewith 1 ≤ k ≤ n.

Denote that S = |x1|+ |x2|+ ... + |xn| = −x1 − x2 − ...− xk + xk+1 + ... + xn

Suppose that the inequality is not true, thus S2n−1 > max{xk+1, ..., xn} and especially

S

2n− 1> −x1 − x2 − ...− xn = S − 2(xk+1 + ... + xn)

Hence, S

2n− 1> S − 2(n− k)

S

2n− 1or S > (2k − 1)S. This contradiction gives us the

proof.

∇1986. Let x1 ≤ x2 ≤ ... ≤ xn ≤ y1 ≤ y2 ≤ ... ≤ yn. Prove that

(x1 + x2 + ... + xn + y1 + y2 + ... + yn)2 ≥ 4n(x1y1 + x2y2 + ... + xnyn).

19

Proof. Setting Pi(t) = t2 − (xi + yi)t + xiyi = (t− xi)(t− yi), i = 1, 2, ..., n. Observe thatfor t ∈ [xi, yi] then Pi(t) ≤ 0.

Now, notice at the quadratic equation:

F (t) =n∑

i=1

Pi(t) = nt2 − (x1 + ... + xn + y1 + ... + yn)t + (x1y1 + ... + xnyn) = 0

As our observation, we have F (t0) ≤ 0, where t0 = xn+y1

2 . Thus exist real roots ofF (t) = 0, hence its discriminant ∆ = (x1 + ... + xn + y1 + ... + yn)2 − 4n(x1y1 + ... + xnyn)has non-negative value.

∇2002. Let a, b, c be positive numbers such that a + b + c = 1. Prove that

1a

+1b

+1c≥ 25

1 + 48abc

Proof. We shall prove the equivalent inequality

1a

+1b

+1c

+ 48 (ab + bc + ca) ≥ 25

W.L.O.G, we may suppose that a+ b ≤ 13√

3. Indeed, if a+ b >

13√

3, b+ c >

13√

3, c+a >

13√

3

then by summing these inequalities we obtain 2 >33√

3, which is false.

Letf (a, b, c) =

1a

+1b

+1c

+ 48 (ab + bc + ca)

We shall prove thatf (a, b, c) ≥ f

(a + b

2,a + b

2, c

)≥ 25

The first inequality in this chain obtains, after simplifications, the form 112

≥ ab(a + b),

which is the consequence of a + b ≤ 13√

3. Denoting a + b

2= l (z = 1 − 2l) in the second

inequality of the chain, after some simplification, we obtain

144l2 − 168l3 + 73l2 − 14l + 1 ≥ 0 ⇔ (3l − 1)2 (4l − 1)2

∇2042. Prove that

tansin x +cotcos x ≥ 2, 0 < x <π

2Proof. Setting sinx = a, cosx = b ⇒ 0 < a, b < 1. The above inequality becomes

(a

b

)a+

(b

a

)b

≥ 2 ⇐⇒ aa+b + ba+b

abba≥ 2

20

By the AM-GM inequality we have

aa+b + ba+b ≥ 2√

(ab)a+b

So we need to prove thata

a−b2 b

b−a2 ≥ 1 ⇐⇒

(a

b

)a−b≥ 1

• If a = b then(a

b

)a−b= 1

• If a > b then a

b> 1, a− b > 0 so

(a

b

)a−b> 1

• If a < b then a

b< 1, a− b < 0 so

(a

b

)a−b> 1

So we are done!

∇

21

Kvant Inequalities

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