Next time, please give opportunity to others to solve the problem alone... That's the idea of this forum.

^He posted it with the "Hide" function, so you don't have to look at it and can solve it by yourself?

I solved it by AM-GM: $\sum_{cyc}{\frac{a}{2b+c^2}}\leq\sum_{cyc}{\frac{a}{3\sqrt[3]{b^2c^2}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{5}{3}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{5}{3}}(abc)^{\frac{1}{9}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{16}{9}}b^{\frac{1}{9}}c^{\frac{1}{9}}}\leq\frac{1}{3}\sum_{cyc}{(\frac{8}{9}a^2+\frac{1}{18}b^2+\frac{1}{18}c^2)}=\frac{a^2+b^2+c^2}{3}$

We can also use tsebyshev and AM-GM to show $$\sum a^2\geq \frac{1}{3}\sum a^{\frac{5}{3}} \sum a^{\frac{1}{3}}\geq \sum a^{\frac{5}{3}}.$$

Also nice way, socrates. I was mistaken when I thought you're wrong, sorry.

Prove that for all positive real numbers $a$, $b$, $c$ such that $abc=1$ the following inequality holds: $$\frac{a^2}{2b+c^2}+\frac{b^2}{2c+a^2}+\frac{c^2}{2a+b^2}\le \frac{a^3+b^3+c^3}3.$$

by AM-GM as math90 do: $\sum_{cyc}{\frac{a^2}{2b+c^2}}\leq\sum_{cyc}{\frac{a^2}{3\sqrt[3]{b^2c^2}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{8}{3}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{8}{3}}(abc)^{\frac{1}{9}}}=\frac{1}{3}\sum_{cyc}{a^{\frac{25}{9}}b^{\frac{1}{9}}c^{\frac{1}{9}}}\leq\frac{1}{3}\sum_{cyc}{(\frac{25}{27}a^3+\frac{1}{27}b^3+\frac{1}{27}c^3)}=\frac{a^3+b^3+c^3}{3}$

$$a^{\frac{2n+1}{n+1}}-a^2\leq -\frac{1}{n+1}(a-1)\leq -\frac{1}{n+1}\ln a\ (n\geq 2)$$ sqing wrote: Prove that for all positive real numbers $a$, $b$, $c$ such that $abc=1$ the following inequality holds: $$\frac{a^2}{2b+c^2}+\frac{b^2}{2c+a^2}+\frac{c^2}{2a+b^2}\le \frac{a^3+b^3+c^3}3.$$ In this case, for $n=2$, after applying the AM-GM inequality to the left hand on the given inequality that others have already shown, hereby, we have $$a^{\frac 53}-a^2\leq -\frac 13(a-1)\leq -\frac 13\ln a.$$

$\frac{a}{2b+c^2} = \frac{a}{b+b+c^2} \le \frac{a}{3\sqrt[3]{b^2c^2}} = \frac{a^{\frac{5}{3}}}{3}$ So we need $\sum a^2 \ge \sum a^{\frac{5}{3}}$, which is AM-GM + Chebyshev

$$\frac{a}{2b+c^2} = \frac{a}{b+b+c^2} \le \frac{a}{3\sqrt[3]{b^2c^2}} = \frac{a^{\frac{5}{3}}}{3}$$$$a^2+b^2+c^2 \ge a^{\frac{5}{3}} + b^{\frac{5}{3}}+ c^{\frac{5}{3}}$$$$(\frac{26}{9},\frac{8}{9},\frac{8}{9} )\ge (\frac{8}{3},1,1)$$$abc=1 $ and by Muirhed inequality $$a^2+b^2+c^2 \ge a^{\frac{5}{3}} + b^{\frac{5}{3}}+ c^{\frac{5}{3}}$$

By AM-GM: $$\sum_{\text{cyc}}\frac a{b+b+c^2}\le\sum_{\text{cyc}}\frac a{3(bc)^{2/3}}=\sum_{\text{cyc}}\frac{a^{5/3}}3$$so it suffices to show $a^2+b^2+c^2\ge a^{5/3}+b^{5/3}+c^{5/3}$. Since $abc=1$, this is equivalent to: $$a^2+b^2+c^2\ge a^{16/9}b^{1/9}c^{1/9}+a^{1/9}b^{16/9}c^{1/9}+a^{1/9}b^{1/9}c^{16/9}$$which is Muirhead $(2,0,0)\succ\left(\frac{16}9,\frac19,\frac19\right)$.