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.

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.
