Inequality $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5} $ I have trouble with solving this inequality:
Prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{3}{5}$ for a,b,c>0.
Using Cauchy-Schwartz I got this: $\frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3} \geq \frac{3}{5} $
Now I have to prove $\frac{a^3}{3ab^2+2c^3} +\frac{b^3}{3bc^2+2a^3} +\frac{c^3}{3ca^2+2b^3} \geq \frac{a^3}{3b^3+c^3} +\frac{b^3}{3c^3+2a^3} +\frac{c^3}{3a^3+2b^3}$
Any help is appreciated!

By Am-Gm inequality we have $$3ab^2\leq a^3+b^3+b^3$$ so $$3ab^2 +2c^3\leq a^3+2b^3+2c^3$$ so  $${a^3 \over 3ab^2+2c^3} \geq {a^3\over a^3+2b^3+2c^3}$$ Let $x=a^3$, $y=b^3$ and $z=c^3$. We can assume that $x+y+z=1$, and so we have to prove $$f(x) +f(y) +f(z)\geq 3/5$$ where $$f(x) ={x\over 2-x}$$ The tangent at $x=1/3$ is $$t(x)={18\over 25}x- {1\over 25}$$ Since $f(x) \geq t(x)$ for all $x\in[0,1)$ we have $$f(x) +f(y) +f(z)\geq t(x) +t(y) +t(z) = 3/5$$
