<s>[INST] 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.
 [/INST] 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.
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