# A four variable inequality Let a,b,c,d be non-negative real numbers satisfying a+b+c+d=3. Show a/(1+2b^3)+b/(1+2c^3)+c/(1+2d^3)+d/(1+2a^3) >= (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).

A four variable inequality
Let $a,b,c,d$ be non-negative real numbers satisfying $a+b+c+d=3$. Show
$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{d}^{3}}+\frac{d}{1+2{a}^{3}}⩾\frac{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}{3}.$
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 : $\left(2,1,0,0\right)$ works (whereas $\left(2,0,1,0\right)$ 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
$\frac{a}{1+2{b}^{3}}+\frac{b}{1+2{c}^{3}}+\frac{c}{1+2{a}^{3}}⩾\frac{{a}^{2}+{b}^{2}+{c}^{2}}{3}$
if $a+b+c=3$. The three variable version itself is implied by the two variable version, but the two variable version is false.
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$\sum _{cyc}\frac{a}{1+2{b}^{3}}=3+\sum _{cyc}\left(\frac{a}{1+2{b}^{3}}-a\right)=3-\sum _{cyc}\frac{2a{b}^{3}}{1+2{b}^{3}}\ge$
$\ge 3-\sum _{cyc}\frac{2a{b}^{3}}{3{b}^{2}}=\frac{\left(a+b+c+d{\right)}^{2}-2\sum _{cyc}ab}{3}=$
$=\frac{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}+ac+bd}{3}\ge \frac{{a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}}{3}.$