Iranian Mathematical Olympiad

Solution 1. By Holder Inequality we have $$\left(\sum _{cyc}\frac{a\sqrt{a}}{bc}\right)\left(\sum _{cyc}bc\right)\left(\sum _{cyc}1\right)\ge {\left(\sum _{cyc}\sqrt{a}\right)}^3\Rightarrow \sum _{cyc}\frac{a\sqrt{a}}{bc}\ge \frac{1}{3}{\left(\sum _{cyc}\sqrt{a}\right)}^3.$$ So it suffices to have $$\frac{1}{3}{\left(\sum _{cyc}\sqrt{a}\right)}^3\ge \sqrt{3}\left(\sum _{cyc}\sqrt{a}\right)\iff {\left(\sum _{cyc}\sqrt{a}\right)}^2\ge 3\sqrt{3}\iff \sum _{cyc}\sqrt{a}\ge \sqrt[4]{27}$$ Now suppose that ${\sum }_{cyc}\sqrt{a}<\sqrt[4]{27}$. It suffices to prove that ${\sum }_{cyc}\frac{a\sqrt{a}}{bc}\ge \sqrt{3\sqrt[4]{27}}$, but we have ${\sum }_{cyc}\frac{a\sqrt{a}}{bc}\ge 3\sqrt[3]{{\prod }_{cyc}\frac{a\sqrt{a}}{bc}}=3(abc{)}^{-\frac{1}{6}}$ (*) by AM-GM inequality $$\begin{array}{rl}1& =\sum _{cyc}ab\ge 3(abc{)}^{\frac{2}{3}}\Rightarrow (abc{)}^{-\frac{2}{3}}\ge 3\Rightarrow (abc{)}^{-\frac{1}{6}}\ge \sqrt[4]{3}\\ \Rightarrow \sum _{cyc}\frac{a\sqrt{a}}{bc}& \ge 3(abc{)}^{-\frac{1}{6}}\ge 3\sqrt[4]{3}=3^{\frac{5}{4}}=3^{\frac{1}{2}}\cdot 3^{\frac{3}{4}}=\sqrt{3\sqrt[4]{27}}.\end{array}$$

□ Solution 2. First by Cauchy-Schwarz inequality we have $$\left(\sum _{cyc}\frac{a^2}{bc\sqrt{a}}\right)\left(\sum _{cyc}bc\sqrt{a}\right)\ge {\left(\sum _{cyc}a\right)}^2.$$ So it suffices to prove that $${\left(\sum _{cyc}a\right)}^2\sqrt{\sum _{cyc}ab}\ge \sqrt{3}\times \sqrt{abc}\left(\sum _{cyc}\sqrt{bc}\right)\left(\sum _{cyc}\sqrt{a}\right).$$ Since ${\sum }_{cyc}ab=1$. But by AM-GM and Cauchy-Schwarz we have $$\sqrt{\sum _{cyc}ab}\ge \frac{1}{\sqrt{3}}\sum _{cyc}\sqrt{ab}(\text{Cauchy-Schwarz})(1)$$ $$\sum _{cyc}a\ge \frac{1}{3}{\left(\sum _{cyc}\sqrt{a}\right)}^2(\text{Cauchy-Schwarz})(2)$$