Problem 1210

Solution 1. Using the given condition $a b c=1$ we get the following:

$$\begin{aligned} \sum_{\text {cyc }} \frac{a}{2 b+c^{2}} & =\sum_{\text {cyc }} \frac{a}{b+b+c^{2}} \\ & \stackrel{\text { AM-GM }}{\leqslant} \sum_{\text {cyc }} \frac{a}{3 \sqrt[3]{b^{2} c^{2}}}=\sum_{\text {cyc }}\left(\frac{a}{3} \cdot \sqrt[3]{a^{2}}\right) \\ & \stackrel{\text { GM-AM }}{\leqslant} \sum_{\text {cyc }}\left(\frac{a}{3} \cdot \frac{a+a+1}{3}\right)=\sum_{\text {cyc }} \frac{a(2 a+1)}{9}=\frac{2}{9} \sum_{\text {cyc }} a^{2}+\frac{1}{9} \sum_{\text {cyc }} a . \end{aligned}$$

Now it suffices to prove that $\frac{2}{9} \sum_{\text {cyc }} a^{2}+\frac{1}{9} \sum_{\text {cyc }} a \leqslant \frac{1}{3} \sum_{\text {cyc }} a^{2}$, which is equivalent with $\sum_{\text {cyc }} a^{2} \geqslant$ $\sum_{\text {cyc }} a$ and that can be easily proven in the following way:

$$\sum_{\mathrm{cyc}} a^{2} \stackrel{\mathrm{QM}-\mathrm{AM}}{\geqslant} 3 \cdot\left(\frac{\sum_{\mathrm{cyc}} a}{3}\right)^{2}=\frac{\sum_{\mathrm{cyc}} a}{3} \cdot \sum_{\mathrm{cyc}} a \stackrel{\mathrm{A}-\mathrm{G}}{\geqslant} \frac{\sum_{\mathrm{cyc}} a}{3} \cdot 3 \sqrt[3]{a b c}=\sum_{\mathrm{cyc}} a$$