Question

$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2$

score 117 · Answer 1 · 2018-01-11 15:07:36

117

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$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2$

Published on 11 Jan 2018 - 15:07

score 1.2k · Answer 2 · 2026-03-26 22:55:38

1.2k

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Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

Published on 26 Mar 2026 - 22:55

score 1.4k · Answer 3 · 2026-03-26 14:20:34

1.4k

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Combined AM GM QM inequality

Published on 26 Mar 2026 - 14:20

score 1.4k · Answer 4 · 2011-08-12 12:46:48

1.4k

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Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$

Published on 12 Aug 2011 - 12:46

score 171 · Answer 5 · 2012-03-06 07:14:49

171

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Find the minimum value of the quantity where $a , b , c$ are real positive numbers.

Published on 06 Mar 2012 - 7:14

score 295 · Answer 6 · 2026-03-26 14:22:18

295

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If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$

Published on 26 Mar 2026 - 14:22

score 310 · Answer 7 · 2012-06-07 22:54:34

310

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Theorem about two real numbers

Published on 07 Jun 2012 - 22:54

score 262 · Answer 8 · 2012-06-08 18:36:35

262

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Maximum product $ab$ for $a+b=k$ - proof based on AM-GM inequality

Published on 08 Jun 2012 - 18:36

score 599 · Answer 9 · 2026-03-26 17:53:50

599

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proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$.

Published on 26 Mar 2026 - 17:53

score 512 · Answer 10 · 2012-07-09 16:54:18

512

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Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$

Published on 09 Jul 2012 - 16:54

score 1.3k · Answer 11 · 2012-07-25 04:33:10

1.3k

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Minimum value of given expression

Published on 25 Jul 2012 - 4:33

score 1.9k · Answer 12 · 2026-03-25 21:49:04

1.9k

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Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$

Published on 25 Mar 2026 - 21:49

score 967 · Answer 13 · 2012-08-31 05:14:24

967

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Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$

Published on 31 Aug 2012 - 5:14

score 397 · Answer 14 · 2026-03-26 22:55:44

397

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Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$

Published on 26 Mar 2026 - 22:55

score 195 · Answer 15 · 2026-03-25 22:26:45

195

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Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum?

Published on 25 Mar 2026 - 22:26

List Question

$\frac{a}{1+b^2c}+\frac{b}{1+c^2d}+\frac{c}{1+d^2a}+\frac{d}{1+a^2b}\geq 2$

Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

Combined AM GM QM inequality

Challenging inequality: $abcde=1$, show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}+\frac{33}{2(a+b+c+d+e)}\ge{\frac{{83}}{10}}$

Find the minimum value of the quantity where $a , b , c$ are real positive numbers.

If $a+b+c+d=16$, then $(a+\frac{1}{c})^2+(c+\frac{1}{a})^2 + (b+\frac{1}{d})^2 + (d+\frac{1}{b})^2 \geq \frac{289}{4}$

Theorem about two real numbers

Maximum product $ab$ for $a+b=k$ - proof based on AM-GM inequality

proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$.

Proving :$\frac{1}{2ab^2+1}+\frac{1}{2bc^2+1}+\frac{1}{2ca^2+1}\ge1$

Minimum value of given expression

Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$

Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$

Inequality. $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a} \geq \frac{ab+bc+ca}{2}$

Does $1 + \frac{1}{x} + \sqrt{\frac{2x}{x + 1}},$ have a global minimum?

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