What's the approach for this inequality question?

Solution 1:

We have \begin{align*} \sum_{\mathrm{cyc}} \frac{a}{\sqrt{a^3 + 5}} &= \sum_{\mathrm{cyc}} \frac{a}{\sqrt{\frac{a^3}{2} + \frac{a^3}{2} + \frac12 + \frac92}}\\ &\le \sum_{\mathrm{cyc}} \frac{a}{\sqrt{3\sqrt[3]{\frac{a^3}{2} \cdot \frac{a^3}{2} \cdot \frac12} + \frac92}}\\ &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 18}}\\ &= \sum_{\mathrm{cyc}} \frac{2a}{\sqrt{6a^2 + 6(ab + bc + ca)}}\\ &= \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} 2\sqrt{\frac{a}{a + b}} \sqrt{\frac{a}{a + c}}\\ &\le \frac{1}{\sqrt{6}}\sum_{\mathrm{cyc}} \left(\frac{a}{a + b} + \frac{a}{a + c}\right)\\ &= \frac{\sqrt{6}}{2}. \end{align*}

We are done.