Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$ 
Let $a$, $b$ and $c$ be positive real numbers. Prove that: 
  $$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$

My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, we have to prove that
$\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2} \geq \frac{3}{5}$
By C-S,
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(2y^2x+3y^2z^2)\frac{x}{z}\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(x^2y^2+3y^2xz\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$

Let $a = x^2, b = y^2, c=z^2$, where $x, y, z > 0$. By C-S, we have 
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \sum_{cyc} \frac{2yx + 3z^2}{zx} \geq \left( \sum_{cyc} \frac{x}{y} \right)^2.$$
But 
$$ \sum_{cyc} \frac{2yx + 3z^2}{zx} = 5 \sum_{cyc} \frac{x}{y}.$$
Combining them we get
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \geq \frac{1}{5} \sum_{cyc} \frac{x}{y}.$$
It remains to apply AM-GM.
