Inequality $\sum_{cyc}a^3\frac{(16a^2-10ab+12b^2)}{(13a^2+5b^2)^2}\geq \frac{a+b+c}{18}$

Question

It's a variant of Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$ :

Let $a,b,c>0$ then we have : $$\sum_{cyc}a^3\frac{(16a^2-10ab+12b^2)}{(13a^2+5b^2)^2}\geq \frac{a+b+c}{18}$$

My proof

Rewriting the inequality we have : $$\sum_{cyc}\frac{a}{a+b+c}\frac{(16-10\frac{b}{a}+12(\frac{b}{a})^2)}{(13+5(\frac{b}{a})^2)^2}\geq \frac{1}{18}$$

Since the function $f(x)=\frac{16-10x+12x^2}{(13+5x^2)^2}$ is convex on $[0.02,+\infty[$ we can apply Jensen's inequality we get :

$$\frac{a}{a+b+c}f(\frac{b}{a})+\frac{b}{a+b+c}f(\frac{c}{b})+\frac{c}{a+b+c}f(\frac{a}{c})\geq f\Big(\frac{a\frac{b}{a}+b\frac{c}{b}+c\frac{a}{c}}{a+b+c}\Big)=\frac{1}{18}$$

So we prove the inequality for $\frac{b}{a},\frac{c}{b},\frac{a}{c}\in[0.02,+\infty]$

The rest is smooth.

My question have you an alternative proof ?

Ps: I will (i have not the time today) make an update with strong convexity .

Answer 1

$$\sum_{cyc}\frac{a^3(16a^2-10ab+12b^2)}{(13a^2+5b^2)^2}-\frac{a+b+c}{18}=$$ $$=\sum_{cyc}\left(\frac{a^3(16a^2-10ab+12b^2)}{(13a^2+5b^2)^2}-\frac{a}{18}-\frac{a-b}{54}\right)=$$ $$=\sum_{cyc}\frac{(a-b)^2(188a^3+5a^2b-50ab^2+25b^3)}{54(13a^2+5b^2)^2}\geq0$$ because by AM-GM: $$188a^3+5a^2b-50ab^2+25b^3=188a^3+5a^2b+3\cdot\frac{25b^3}{3}-50ab^2\geq$$ $$\geq\left(5\sqrt[5]{188\cdot5\cdot\left(\frac{25}{3}\right)^3}-50\right)ab^2\geq0.$$