Question

Duality gap in polynomial optimization problem Lasserre relaxation

score 145 · Answer 1 · 2026-03-17 06:37:03

145

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Duality gap in polynomial optimization problem Lasserre relaxation

Published on 17 Mar 2026 - 6:37

score 531 · Answer 2 · 2026-03-24 03:09:54

531

Views

$\;32\displaystyle\left(\sum_{c}\frac{1}{7+(x-3)^2}\right)\leq \sum_{c} \frac{x^2+yz}{y+z}+6$

Published on 24 Mar 2026 - 3:09

score 919 · Answer 3 · 2026-03-25 01:37:10

919

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Prove that $(x−2y+z)^2 \geq 4xz−8y$

Published on 25 Mar 2026 - 1:37

score 92 · Answer 4 · 2026-03-24 18:46:22

92

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Prove that the inequality is true

Published on 24 Mar 2026 - 18:46

score 131 · Answer 5 · 2018-12-03 13:29:53

131

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Inequality for $a,c,b$ positive real numbers with $a + b + c = 1$

Published on 03 Dec 2018 - 13:29

score 123 · Answer 6 · 2018-12-09 15:42:37

Prove that $\frac{3a^3+7b^3}{2a+3b}+\frac{3b^3+7c^3}{2b+3c}+\frac{3c^3+7a^3}{2c+3a}\ge 3\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)$

score 199 · Answer 7 · 2018-12-16 12:14:19

199

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A triangle has sides $a$, $b$, $c$ and medians $m_a$, $m_b$, $m_c$. Show $(ab+bc+ca)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 2\sqrt{3}(m_a+m_b+m_c)$

Published on 16 Dec 2018 - 12:14

score 149 · Answer 8 · 2018-12-18 12:58:48

149

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If $ab+bc+ca=3$ for non-negative $a$, $b$, $c$, show that $\sum_{cyc}a^2b^2+\sum_{cyc}\frac{12a^2b^2c^2}{(a+b)^2}\ge 12abc$

Published on 18 Dec 2018 - 12:58

score 137 · Answer 9 · 2018-12-19 06:29:34

137

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For $x$, $y$, $z$ the sides of a triangle, show $\sum_{cyc}\frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}\ge\frac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$

Published on 19 Dec 2018 - 6:29

score 53 · Answer 10 · 2026-03-22 07:21:07

53

Views

Inequality triangle Radon substitutions

Published on 22 Mar 2026 - 7:21

score 215 · Answer 11 · 2019-01-25 02:43:56

215

Views

Inequality. ${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$

Published on 25 Jan 2019 - 2:43

score 130 · Answer 12 · 2026-03-25 06:11:00

130

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Inequality with $a+b+c=1$, a,b,c positive numbers

Published on 25 Mar 2026 - 6:11

score 217 · Answer 13 · 2026-03-24 23:46:45

217

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Prove $\sum \sqrt{\frac{a^2}{6a^2+5ab+b^2}}\le \frac{\sqrt{3}}{2}$

Published on 24 Mar 2026 - 23:46

score 238 · Answer 14 · 2019-04-22 07:34:27

238

Views

Decomposing bivariate quadratic form into sum of two squares

Published on 22 Apr 2019 - 7:34

score 295 · Answer 15 · 2026-03-25 06:06:19

295

Views

Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .

Published on 25 Mar 2026 - 6:06

List Question

Duality gap in polynomial optimization problem Lasserre relaxation

$\;32\displaystyle\left(\sum_{c}\frac{1}{7+(x-3)^2}\right)\leq \sum_{c} \frac{x^2+yz}{y+z}+6$

Prove that $(x−2y+z)^2 \geq 4xz−8y$

Prove that the inequality is true

Inequality for $a,c,b$ positive real numbers with $a + b + c = 1$

Prove that $\frac{3a^3+7b^3}{2a+3b}+\frac{3b^3+7c^3}{2b+3c}+\frac{3c^3+7a^3}{2c+3a}\ge 3\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)$

A triangle has sides $a$, $b$, $c$ and medians $m_a$, $m_b$, $m_c$. Show $(ab+bc+ca)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 2\sqrt{3}(m_a+m_b+m_c)$

If $ab+bc+ca=3$ for non-negative $a$, $b$, $c$, show that $\sum_{cyc}a^2b^2+\sum_{cyc}\frac{12a^2b^2c^2}{(a+b)^2}\ge 12abc$

For $x$, $y$, $z$ the sides of a triangle, show $\sum_{cyc}\frac{yz((y+z)^2-x^2)}{(y^2+z^2)^2}\ge\frac{9(y+z-x)(x+z-y)(x+y-z)}{4xyz}$

Inequality triangle Radon substitutions

Inequality. ${{\sqrt{a}+\sqrt{b}+\sqrt{c}} \over {2}} \ge {{1} \over {\sqrt{a}}} + {{1} \over {\sqrt{b}}} + {{1} \over {\sqrt{c}}}$

Inequality with $a+b+c=1$, a,b,c positive numbers

Prove $\sum \sqrt{\frac{a^2}{6a^2+5ab+b^2}}\le \frac{\sqrt{3}}{2}$

Decomposing bivariate quadratic form into sum of two squares

Given three non-negatve numbers $a,b,c$. Prove that $1+a^{2}+b^{2}+c^{2}+4abc\geqq a+b+c+ab+bc+ca$ .

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