List questions from sum-of-squares-method

Published on 25 Mar 2026 - 23:08

140

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Somebody help me please. I have a difficult inequality.

Published on 25 Mar 2026 - 6:27

197

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Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$

Published on 16 Jul 2017 - 19:11

168

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Prove this inequality $\sum_{cyc}\frac{a}{ab+1}\geq \frac{3}{2}$

Published on 30 Jul 2017 - 15:12

91

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Inequality very very long

Published on 02 Aug 2017 - 13:52

166

Views

Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$

Published on 25 Mar 2026 - 17:42

407

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Inequality : $ \sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}$

Published on 12 Aug 2017 - 12:18

388

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Wanted: Low-dimensional SOS certificate for the AM-GM inequality

Published on 18 Aug 2017 - 9:49

178

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If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$

Published on 25 Mar 2026 - 20:22

145

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Prove this inequality $\sum _{cyc}a^3+4\left(\sum _{cyc}\frac{ab}{a^2+b^2}\right)\ge 9$

Published on 23 Aug 2017 - 6:48

208

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Is it possible to prove that for $x,y,z \in \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $(x+y+z)\ge 3$ without using the AM-GM inequality?

Published on 25 Aug 2017 - 18:15

333

Views

Inequality with $ab + bc + ca = 1$

Published on 26 Aug 2017 - 2:12

344

Views

How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Published on 25 Mar 2026 - 20:22

6.1k

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Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms

Published on 30 Aug 2017 - 11:52

456

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Turkevicius inequality

Published on 25 Mar 2026 - 19:01

List Question

Prove the inequality: $a^2+b^2+1≥ab+a+b$

Somebody help me please. I have a difficult inequality.

Find minimum of $a+b+c+\frac1a+\frac1b+\frac1c$ given that: $a+b+c\le \frac32$

Prove this inequality $\sum_{cyc}\frac{a}{ab+1}\geq \frac{3}{2}$

Inequality very very long

Prove the inequality $\sum_{cyc}\frac{a}{1+\left(b+c\right)^2}\le \frac{3\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2+12abc}$

Inequality : $ \sum_{cyc} \frac{a+b}{\sqrt{a+2c}} \geq 2 \sqrt{a+b+c}$

Wanted: Low-dimensional SOS certificate for the AM-GM inequality

If a, b, c>0 show that $\frac{a^2}{b^2+bc}+\frac{b^2}{c^2+ac}+\frac{c^2}{a^2+ab} \ge \frac{3}{2}$

Prove this inequality $\sum _{cyc}a^3+4\left(\sum _{cyc}\frac{ab}{a^2+b^2}\right)\ge 9$

Is it possible to prove that for $x,y,z \in \Bbb{R}$; if $x,y,z >0$ and $xyz=1$, then $(x+y+z)\ge 3$ without using the AM-GM inequality?

Inequality with $ab + bc + ca = 1$

How prove this inequality $x^3y+y^3z+z^3x\ge xyz(x+y+z)$

Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms

Turkevicius inequality

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