A four variable inequality

Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show

$$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{d}^{3}}+\frac{d}{1+2{a}^{3}}\u2a7e\frac{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}{3}.$$

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

$$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{a}^{3}}\u2a7e\frac{{a}^{2}+{b}^{2}+{c}^{2}}{3}$$

if $a+b+c=3$. The three variable version itself is implied by the two variable version, but the two variable version is false.

Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show

$$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{d}^{3}}+\frac{d}{1+2{a}^{3}}\u2a7e\frac{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}{3}.$$

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

$$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{a}^{3}}\u2a7e\frac{{a}^{2}+{b}^{2}+{c}^{2}}{3}$$

if $a+b+c=3$. The three variable version itself is implied by the two variable version, but the two variable version is false.