Question

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

score 123 · Answer 1 · 2026-03-31 12:00:20

123

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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 31 Mar 2026 - 12:00

score 310 · Answer 2 · 2026-05-10 18:42:25

310

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Inequality with $a+b+c=1$

Published on 10 May 2026 - 18:42

score 158 · Answer 3 · 2026-03-25 22:05:20

158

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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 25 Mar 2026 - 22:05

score 157 · Answer 4 · 2026-03-31 11:10:59

157

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 31 Mar 2026 - 11:10

score 288 · Answer 5 · 2026-03-31 12:01:29

288

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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 31 Mar 2026 - 12:01

score 164 · Answer 6 · 2026-03-31 07:08:39

164

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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 31 Mar 2026 - 7:08

score 205 · Answer 7 · 2026-03-31 06:31:13

205

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

Published on 31 Mar 2026 - 6:31

score 90 · Answer 8 · 2026-05-11 00:43:26

90

Views

Non - symetric inequality in 3 variables

Published on 11 May 2026 - 0:43

score 1.1k · Answer 9 · 2026-03-30 20:41:53

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 30 Mar 2026 - 20:41

score 181 · Answer 10 · 2026-03-31 13:13:37

181

Views

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

Published on 31 Mar 2026 - 13:13

score 293 · Answer 11 · 2026-03-31 14:55:28

293

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 31 Mar 2026 - 14:55

score 315 · Answer 12 · 2026-03-31 07:00:22

315

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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 31 Mar 2026 - 7:00

score 153 · Answer 13 · 2026-03-30 23:33:37

153

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

Published on 30 Mar 2026 - 23:33

score 114 · Answer 14 · 2026-03-31 11:06:39

114

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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 31 Mar 2026 - 11:06

score 367 · Answer 15 · 2026-03-31 12:31:30

367

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 31 Mar 2026 - 12:31

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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