Question

How prove this inequality with $\sum\sqrt{a+b}\sum\frac{1}{\sqrt{2a+b+c}}\le\frac{9}{\sqrt{2}}$?

score 121 · Answer 1 · 2016-08-20 08:43:31

121

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How prove this inequality with $\sum\sqrt{a+b}\sum\frac{1}{\sqrt{2a+b+c}}\le\frac{9}{\sqrt{2}}$?

Published on 20 Aug 2016 - 8:43

score 307 · Answer 2 · 2026-03-25 09:23:34

307

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Inequality with $a+b+c=1$

Published on 25 Mar 2026 - 9:23

score 157 · Answer 3 · 2026-03-24 03:26:43

157

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Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

Published on 24 Mar 2026 - 3:26

score 155 · Answer 4 · 2016-09-20 17:45:23

155

Views

Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$

Published on 20 Sep 2016 - 17:45

score 286 · Answer 5 · 2016-09-26 12:17:28

286

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I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$

Published on 26 Sep 2016 - 12:17

score 162 · Answer 6 · 2016-10-07 10:23:41

162

Views

Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$

Published on 07 Oct 2016 - 10:23

score 203 · Answer 7 · 2016-10-13 04:17:54

203

Views

if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$

Published on 13 Oct 2016 - 4:17

score 88 · Answer 8 · 2026-03-25 11:11:47

88

Views

Non - symetric inequality in 3 variables

Published on 25 Mar 2026 - 11:11

score 1.1k · Answer 9 · 2016-10-28 07:45:57

1.1k

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Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $

Published on 28 Oct 2016 - 7:45

score 179 · Answer 10 · 2016-11-02 22:19:25

179

Views

Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$

Published on 02 Nov 2016 - 22:19

score 291 · Answer 11 · 2016-12-28 21:32:01

291

Views

Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$

Published on 28 Dec 2016 - 21:32

score 313 · Answer 12 · 2017-01-13 07:26:04

313

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Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$.

Published on 13 Jan 2017 - 7:26

score 151 · Answer 13 · 2017-01-27 00:51:11

151

Views

Prove $(x+y+z)^8\ge 3^7(x^2+y^2+z^2)x^2y^2z^2$

Published on 27 Jan 2017 - 0:51

score 112 · Answer 14 · 2017-02-12 16:26:17

112

Views

Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$

Published on 12 Feb 2017 - 16:26

score 365 · Answer 15 · 2017-02-13 09:28:20

365

Views

Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$

Published on 13 Feb 2017 - 9:28

List Question

How prove this inequality with $\sum\sqrt{a+b}\sum\frac{1}{\sqrt{2a+b+c}}\le\frac{9}{\sqrt{2}}$?

Inequality with $a+b+c=1$

Given $abc=1$ and $0< c \leq b \leq1\leq a$, prove that $8(a+b+c)^2\le9(1+a^2)(1+b^2)(1+c^2)$

Prove the inequality $\frac a{a+b^2}+\frac b{b+c^2}+\frac c{c+a^2}\le\frac14\left(\frac1a+\frac1b+\frac1c\right)$

I conjecture this inequality $\sqrt[4]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}}\ge\sqrt[3]{\frac{(x+y)(y+z)(z+x)}{8}}$

Prove or disprove $\sqrt[3]{\frac{(ab+bc+ac)(a^2+b^2+c^2)}{9}}\ge\sqrt[4]{\frac{(a^2b^2+b^2c^2+c^2a^2)}{3}}$

if $abc=1$,show that $a^3+b^3+c^3+\frac{256}{(a+1)(b+1)(c+1)}\ge 35$

Non - symetric inequality in 3 variables

Prove inequality $ \frac{(x+y-1)^2}{z}+\frac{(x+z-1)^2}{y}+\frac{(y+z-1)^2}{x} \geq 12 $

Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$

Prove that $\sum\limits_{cyc}\frac{a}{a+b}\geq1+\frac{3\sqrt[3]{a^2b^2c^2}}{2(ab+ac+bc)}$

Find the maximum value of the expression ${\frac {x}{1+x^2}} + {\frac {y}{1+y^2}}+{\frac {z}{1+z^2}}$.

Prove $(x+y+z)^8\ge 3^7(x^2+y^2+z^2)x^2y^2z^2$

Minimize $\big(3+2a^2\big)\big(3+2b^2\big)\big(3+2c^2\big)$ if $a+b+c=3$ and $(a,b,c) > 0$

Inequality trouble: $(a^2+ab+b^2)(b^2+bc+c^2)(c^2+ca+a^2) \geq (ab+bc+ca)^3$

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