Prove that $\sum\limits_{cyc}\frac{a^2}{a^3+2}\leq\frac{4}{3}$ if $a, b, c, d > 0$ and $abcd=1$ 
Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that:
$$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}.$$

Vasc's LCF Theorem does not help here. Also I tried MV method, but without success.

Let : $a= e^{x_1}\ , \ b=e^{x_2}\ , \ c=e^{x_3} \ , \ d =e^{x_4}$
So we must to prove:
$$f(x_1)+f(x_2)+f(x_3)+f(x_4) \le \dfrac{4}{3}$$ 
for $x_1+x_2+x_3+x_4=0 $
$$f(x)= \dfrac{e^{2x}}{e^{3x}+2}$$
Since : $f'(x)= \dfrac{e^{2x}(4-e^{3x})}{(2+e^{3x})^2}$ and $f''(x)=\dfrac{e^{2x}(e^{6x}-26e^{3x}+16)}{(2+e^{3x})^3}$
We only need to consider the inequality in case : $x_1=x_2=x_3=t\ , \ x_4= -4t$
$\Leftrightarrow a=b=c=x \ ,\ d=\dfrac{1}{x^3}$
$g(x)=\dfrac{3x^2}{x^3+2}+\dfrac{x^3}{1+2x^9} \ , x>0$
$g'(x)=\dfrac{3x(1-x^4)(4x^{17}-16x^{14}+4x^{13}+4x^{12}-16x^{10}+20x^9+8x^8+4x^5+8x^4-x^3+4x+4)}{(x^3+2)^2(2x^9+1)^2}$
Maximum is attained at $x=1\ ,\ g(1)=\dfrac{4}{3}$
Equality holdes for : $(a=b=c=d=1)$
