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!
 A: 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 $$f(x) +f(y) +f(z)\geq t(x) +t(y) +t(z) = 3/5$$
A: By the Cauchy-Schwartz inequality, we have:
$$(\frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3})(a(3ab^2+2c^3)+b(3bc^2+2a^3)+c(3ca^2+2b^3)) \\ \geq (a^2+b^2+c^2)^2 \\ \implies \frac{a^3}{3ab^2+2c^3}+\frac{b^3}{3bc^2+2a^3}+\frac{c^3}{3ca^2+2b^3} \geq \frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)}.$$
Therefore, we are done if we show that:
$$\frac{(a^2+b^2+c^2)^2}{3(a^2b^2+a^2c^2+b^2c^2)+2(ac^3+ba^3+cb^3)} \geq \frac{3}{5},$$
or equivalently:
$$5(a^4+b^4+c^4)+(a^2b^2+a^2c^2+b^2c^2)\geq 6(ac^3+ba^3+cb^3).$$
However, we already have: $$a^4+a^2b^2 \geq 2ba^3 \\b^4+b^2c^2 \geq 2cb^3 \\c^4+a^2c^2 \geq 2ac^3 .$$
Thus, it is enough to show that:
$$a^4+b^4+c^4 \geq ac^3+ba^3+cb^3,$$
which is a consequence of the rearrangement inequality (note that $a,b,c$ and $a^3,b^3,c^3$ are sorted in the same way).
