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Transcript of 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
21692. 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
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
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
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.
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∇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
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∇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!
∇
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