Solve the inequality $\frac{a^3}{2b^3+ab^2}+\frac{b^3}{2c^3+bc^2}+\frac{c^3}{2d^3+cd^2}+\frac{d^3}{2a^3+da^2}\geq \frac{4}{3}$ for $a,b,c>0$.
I tried to use the inequality $a^3+2b^3\geq 3ab^2$, etc., adding $1$ to each fraction. I tried with Hölder too, but nothing works. Please help me.
Study the function $f(x) = \frac{x^3}{x+2} $ with $x>0$. This function is convex because $f''(x) =\frac{2x(x^2+6x+12)}{(x+2)^3}>0$. Then appling the Jensen's inequality:
$$\frac{1}{4}f\left(\frac{a}{b}\right)+\frac{1}{4}f\left(\frac{b}{c}\right)+\frac{1}{4}f\left(\frac{c}{d}\right)+\frac{1}{4}f\left(\frac{d}{a}\right) \ge f \left(\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+\frac{d}{a}\right)\right) \tag{1}$$
It's easy to notice that
$$\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+ \frac{d}{a}\right) \ge 1 \tag{2}$$
and as $f(x)$ is increasing because $f'(x) = \frac{2x^2(x+3)}{(x+2)^2}>0$ then from $(2)$, we have
$$f \left(\frac{1}{4}\left(\frac{a}{b}+\frac{b}{c}+ \frac{c}{d}+\frac{d}{a}\right)\right) \ge f(1) =\frac{1}{3} \tag{3}$$
From $(1)$ and $(3)$ we conclude that
$$\frac{a^3}{2b^3+ab^2}+\frac{b^3}{2c^3+bc^2}+\frac{c^3}{2d^3+cd^2}+\frac{d^3}{2a^3+da^2}\geq \frac{4}{3}$$
The equality occurs when $a = b= c = d$.