BdMO Online Forum

Let $a \ge b \ge c > 0$. Then by applying chebyshev, we get,
\[ \sum_{cyc}^{} \dfrac{a\sqrt{a}}{bc} \ge \dfrac{1}{3} \left( \sum_{cyc}^{} \dfrac{a}{bc} \right)\left( \sum_{cyc}^{} \sqrt{a} \right) \]

Then it's enough to prove that
\begin{align*}
\dfrac{1}{3} \left( \sum_{cyc} \dfrac{a}{bc} \right) &\ge \sqrt{3} \\
or, \sum a^2 &\ge 3\sqrt{3} abc \\
or, \left( \sum a^2 \right) \left(\sqrt{\sum ab} \right) &\ge 3\sqrt{3} abc \\
\end{align*}

Which is clearly true by AM-GM and we are done