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

Question

Let $a;b;c>0$. Prove that $$\frac{3a^3+7b^3}{2a+3b}+\frac{3b^3+7c^3}{2b+3c}+\frac{3c^3+7a^3}{2c+3a}\ge 3\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)$$

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My proof: That inequality is equivalent

$2a^4b+2b^4c+2c^4a+3ab^4+3bc^4+3ca^4\ge 5a^2b^2c+5a^2bc^2+5ab^2c^2$

It is easy to prove $ab^4+bc^4+ca^4\ge a^2b^2c+a^2bc^2+ab^2c^2$ ( divide both sides by $abc$ and use Cauchy-Schwarz's inequality)

So we need to prove $2a^4b+2b^4c+2c^4a+2ab^4+2bc^4+2ca^4\ge 4a^2b^2c+4a^2bc^2+4ab^2c^2$

$\Leftrightarrow \sum_{cyc} (a-b)^2(2c^3+bc^2-b^2c+ac^2-a^2c+3ab^2+3a^2b)\ge 0$ .Done

This is my proof but I spend a lot of time getting the last inequality. I need another solution use SOS which is easier than mine.

Answer 1

Yes, SOS helps.

Indeed, $$\sum_{cyc}\frac{3a^3+7b^3}{2a+3b}-\sum_{cyc}(3a^2-ab)=\sum_{cyc}\left(\frac{3a^3+7b^3}{2a+3b}-3b^2+ab\right)=$$ $$=\sum_{cyc}\frac{(a^2-b^2)(3a+2b)}{2a+3b}=\sum_{cyc}\left(\frac{(a^2-b^2)(3a+2b)}{2a+3b}-(a^2-b^2)\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2(a+b)}{2a+3b}\geq0.$$