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.
 A: Here is a solution that relies on an uninspiring chain of single-variable inequalities of the form $\frac{x^2}{x^3+2}\leq \alpha+\beta\ln(x)$ on some interval for various nonnegative real $\alpha$ and $\beta$.
Define $f(x)=\frac{x^2}{x^3+2}$. Note first that $f(x)$ is maximized when
$$f'(x)=\frac{2x(x^3+2)-3x^2(x^2)}{(x^3+2)^2}=\frac{x(4-x^3)}{(x^3+2)^2}=0,$$
i.e. when $x=2^{2/3}$, at which point $f(x)=\frac{2^{1/3}}3$. If $$\min(f(a),f(b),f(c),f(d))\leq\frac43-2^{1/3},$$
then without loss of generality let $f(a)$ be less than this bound; then
$$f(a)+f(b)+f(c)+f(d)\leq \left(\frac43-2^{1/3}\right)+\frac{2^{1/3}}3+\frac{2^{1/3}}3+\frac{2^{1/3}}3=\frac43.$$
So, we may assume that
$$f(a),f(b),f(c),f(d)\geq \frac43-2^{1/3}>0.07;$$
this implies $a,b,c,d>0.37$.
Now, define $g(x)=\frac13(1+\ln(x))$. One may check that $f(x)\leq g(x)$ for all $x>0.6$. So, if $a,b,c,d>0.6$, then
$$f(a)+f(b)+f(c)+f(d)\leq \frac43+\frac13\ln(abcd)=\frac43;$$
otherwise, without loss of generality let $a<0.6$; now $0.37<a<0.6$.
Now, define $h(x)=0.36+0.17\ln(x)$; note that $f(x)<h(x)$ for all $x>0.13$. We thus have
\begin{align*}
f(a)+f(b)+f(c)+f(d)&\leq f(a)+1.08+0.17\ln(bcd)\\
&=1.08+\frac{a^2}{a^3+2}-0.17\ln(a). 
\end{align*}
One may verify that the one-variable function
$$j(a)=1.08+\frac{a^2}{a^3+2}-0.17\ln(a)$$
is maximized on the interval $[0.37,0.6]$ at $a=0.6$, where its value is $1.3293<\frac 43$. The inequality is thus proven.
A: WLOG, assume that $a\ge b \ge c \ge d$.
Let
$$f(u) := \frac{u^2}{u^3 + 2}.$$
Fact 1: $f(u)$ is strictly increasing on $(0, \sqrt[3]{4}]$
and strictly decreasing on $[\sqrt[3]{4}, \infty)$.
We split into four cases.
Case 1: $c > \sqrt[3]{13 - 3\sqrt{17}}$
and $a < \sqrt[3]{13 + 3\sqrt{17}}$
Let
$$g(u) := \frac{\mathrm{e}^{2u}}{\mathrm{e}^{3u} + 2}.$$
We have
$$g''(u) = \frac{\mathrm{e}^{2u}(\mathrm{e}^{6u} - 26\mathrm{e}^{3u} + 16)}{(\mathrm{e}^{3u}  + 2)^3}.$$
We have
$g''(u) \le 0$ on $[\ln \sqrt[3]{13 - 3\sqrt{17}}, ~ \ln \sqrt[3]{13 + 3\sqrt{17}}]$.
Thus, using Jensen's inequality, we have
\begin{align*}
 f(a) + f(b) + f(c) &= g(\ln a) + g(\ln b) + g(\ln c)\\
 &\le 3 \, g\left(\frac{\ln a + \ln b + \ln c}{3}\right)\\
 &= 3 \cdot \frac{(abc)^{2/3}}{abc + 2}.
\end{align*}
It suffices to prove that
\begin{align*}
 3 \cdot \frac{(abc)^{2/3}}{abc + 2} + \frac{d^2}{d^3 + 2} \le \frac43
\end{align*}
or
$$3 \cdot \frac{(1/d)^{2/3}}{1/d + 2} + \frac{d^2}{d^3 + 2} \le \frac43 $$
which is true for all $d > 0$. (Note: Let $d = x^3$.)
Case 2: $d < 3/5$ and $c \le \sqrt[3]{13 - 3\sqrt {17}}$
Using Fact 1, we have
$$f(a) + f(b) + f(c) + f(d)
\le f(\sqrt[3]{4}) + f(\sqrt[3]{4})
+ f(\sqrt[3]{13 - 3\sqrt{17}}) + f(3/5) < \frac43.$$
Case 3: $d < 3/5$ and $a \ge \sqrt[3]{13 + 3\sqrt{17}}$
Using Fact 1, we have
$$f(a) + f(b) + f(c) + f(d)
\le f(\sqrt[3]{13 + 3\sqrt{17}}) + f(\sqrt[3]{4})
+ f(\sqrt[3]{4}) + f(3/5) < \frac43.$$
Case 4: $d \ge \frac35$
Fact 2:
$\frac{1}{3} - \frac{u^2}{u^3 + 2} + \frac13 \ln u \ge 0$ for all $u \ge 3/5$.
(Note: Take derivative.)
Using Fact 2, we have
$$\sum_{\mathrm{cyc}} \left(\frac{1}{3} - \frac{a^2}{a^3 + 2} + \frac13 \ln a\right) \ge 0.$$
The desired result follows.
We are done.
A: 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 : $(a=b=c=d=1)$
