a four variable inequality Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show
\begin{equation*}
\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2d^3}+\frac{d}{1+2a^3} \geqslant \frac{a^2+b^2+c^2+d^2}{3}.
\end{equation*}
I got this inequality from an IMO preparation club.  I no longer do contest math  but still I find this interesting. There is a weird equality case : $(2,1,0,0)$ works (whereas $(2,0,1,0)$ does not).
So far I've tried regrouping terms to disminish the number of variables or the number of non-zero variables. I've shown that the inequality is implied by the three-variable version
\begin{equation*}
\frac{a}{1+2b^3}+\frac{b}{1+2c^3}+\frac{c}{1+2a^3} \geqslant \frac{a^2+b^2+c^2}{3}
\end{equation*} if $a+b+c=3$. The three variable version itself is implied by the two variable version, but the two variable version is false.

By AM-GM $$\sum_{cyc}\frac{a}{1+2b^3}=3+\sum_{cyc}\left(\frac{a}{1+2b^3}-a\right)=3-\sum_{cyc}\frac{2ab^3}{1+2b^3}\geq$$
$$\geq3-\sum_{cyc}\frac{2ab^3}{3b^2}=\frac{(a+b+c+d)^2-2\sum\limits_{cyc}ab}{3}=$$
$$=\frac{a^2+b^2+c^2+d^2+ac+bd}{3}\geq\frac{a^2+b^2+c^2+d^2}{3}.$$
