Question

$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

score 79 · Answer 1 · 2026-03-25 13:56:25

79

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$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

Published on 25 Mar 2026 - 13:56

score 156 · Answer 2 · 2020-04-25 08:12:06

156

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Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $

Published on 25 Apr 2020 - 8:12

score 81 · Answer 3 · 2026-03-26 11:16:47

81

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Does this inequality hold with some constant factor $c>0$?

Published on 26 Mar 2026 - 11:16

score 40 · Answer 4 · 2026-03-26 11:16:21

40

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Is this "constrained" infimum zero?

Published on 26 Mar 2026 - 11:16

score 130 · Answer 5 · 2020-04-28 16:47:37

130

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Prove that $a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$

Published on 28 Apr 2020 - 16:47

score 307 · Answer 6 · 2026-03-27 12:50:26

307

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MOP 2011 inequality

Published on 27 Mar 2026 - 12:50

score 140 · Answer 7 · 2026-03-25 13:56:19

140

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If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

Published on 25 Mar 2026 - 13:56

score 92 · Answer 8 · 2020-05-01 13:59:05

92

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

Published on 01 May 2020 - 13:59

score 69 · Answer 9 · 2020-05-03 06:28:35

69

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Proof that $\sum a_k a_{k+1} \leq \dfrac{a^2}{4} $ where $a = \sum a_k $

Published on 03 May 2020 - 6:28

score 1.1k · Answer 10 · 2020-05-03 09:27:44

1.1k

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Is the AM-GM inequality the only obstruction for getting a specific sum and product?

Published on 03 May 2020 - 9:27

score 150 · Answer 11 · 2020-05-05 09:19:11

150

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Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$

Published on 05 May 2020 - 9:19

score 134 · Answer 12 · 2026-03-25 19:07:07

134

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If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $

Published on 25 Mar 2026 - 19:07

score 187 · Answer 13 · 2026-03-25 16:20:52

187

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If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $

Published on 25 Mar 2026 - 16:20

score 35 · Answer 14 · 2020-05-08 05:13:07

35

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finding x,y,z for optimizing expression

Published on 08 May 2020 - 5:13

score 153 · Answer 15 · 2020-05-08 09:37:44

153

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Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$

Published on 08 May 2020 - 9:37

List Question

$ \frac{a^{2}}{a+2 b^{3}}+\frac{b^{2}}{b+2 c^{3}}+\frac{c^{2}}{c+2 a^{3}} \geq 1 $

Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $

Does this inequality hold with some constant factor $c>0$?

Is this "constrained" infimum zero?

Prove that $a^{2} \sqrt{b^{2}-b c+c^{2}}+b^{2} \sqrt{c^{2}-c a+a^{2}}+c^{2} \sqrt{a^{2}-a b+b^{2}} \leq a^{3}+b^{3}+c^{3}$

MOP 2011 inequality

If $a+b+c=3$ Prove that $a^{2}+b^{2}+c^{2}\geq\frac{2+a}{2+b}+\frac{2+b}{2+c}+\frac{2+c}{2+a}$

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

Proof that $\sum a_k a_{k+1} \leq \dfrac{a^2}{4} $ where $a = \sum a_k $

Is the AM-GM inequality the only obstruction for getting a specific sum and product?

Prove that $\sqrt{\frac{2 a^{2}+b c}{a^{2}+2 b c}}+\sqrt{\frac{2 b^{2}+c a}{b^{2}+2 c a}}+\sqrt{\frac{2 c^{2}+a b}{c^{2}+2 a b}} \geq 2 \sqrt{2}$

If $x+y+z=1$ prove $ \sqrt{x+\frac{(y-z)^{2}}{12}}+\sqrt{y+\frac{(z-x)^{2}}{12}}+\sqrt{z+\frac{(x-y)^{2}}{12}} \leq \sqrt{3} $

If $a+b+c+d=4$ Prove that $ \sqrt{\frac{a+1}{a b+1}}+\sqrt{\frac{b+1}{b c+1}}+\sqrt{\frac{c+1}{c d+1}}+\sqrt{\frac{d+1}{d a+1}} \geq 4 $

finding x,y,z for optimizing expression

Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$

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