# Questions tagged [sum-of-squares-method]

Proofs of inequalities by the Sum of Squares method (SOS).

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### Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$

Reducing this whole expression i finally came to this $$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$ Here I am stuck. I can't prove this. ...
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### Prove $P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$

For $a,b,c$ are reals$.$ Prove$:$ $$P= 7\,{c}^{4}-2\,ab{c}^{2}-2\,ab \left( a+b \right) c+ \left( a+b \right) ^{2} \left( {a}^{2}+{b}^{2} \right) \geqq 0$$ I found this from Michael Rozenberg's ...
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### Prove that $\sum_{\mathrm{cyc}} (40a^6 + 53a^5b) \ge 0$

(P1) Let $a, b, c$ be real numbers. Prove that $40(a^6+b^6+c^6) + 53(a^5b+b^5c+c^5a) \ge 0.$ This inequality is verified by Mathematica. I am particularly interested in (simple) SOS solutions (also ...
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### Any good usable software or software package for decomposition with constraint?

Does anyone know any good and usable software package, preferably in Windows, that can effectively find SOS (sum of squares) polynomials $s_{0}\left(x\right)$ and $s_{1}\left(x\right)$ for any real ...
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### Prove $\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$

For $a,\,b,\,c>0$. Prove: $$\frac{3}{2} +\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \leqq \frac{a}{b}+\frac{b}{c} +\frac{c}{a}$$ My work: After a lot of caculates, I found: $\text{RHS-LHS}=$ ...
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For $a,b,c > 0$ prove: $$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}) +\frac{486(ab+bc+ca)^3}{(a+b+c)^6} \geqq 27$$ My work: I can easy found SOS for it: $$\text{LHS-RHS}=\sum {\... 3answers 114 views ### How can I approach this inequality? [duplicate] Let a, b and c be three non-zero positive numbers. Show that:$$\sqrt{\frac{2a}{a + b}} + \sqrt{\frac{2b}{b + c}} + \sqrt{\frac{2c}{a + c}} \leq 3$$I know the triangular inequality would help ... 3answers 84 views ### show this inequality \sum_{cyc}\frac{1}{5-2xy}\le 1 let x,y,z\ge 0 and such x^2+y^2+z^2=3 show that$$\sum_{cyc}\dfrac{1}{5-2xy}\le 1$$try:$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le 2$$and$$\sum_{cyc}\dfrac{2xy}{5-2xy}\le\sum_{cyc}\dfrac{(x+y)^2}{\frac{...
Consider the following quadratic function \begin{equation} f(y,z)=1697 y^2+57 y z+81 y+407 z^2-6 z+1 \end{equation} Using first and second derivatives test, we have shown that the global minimum of $f$...