The 14th Thailand Mathematical Olympiad

Assume without loss of generality that $a^2+b^2=c^2$. By the AM-GM inequality, $$c^2=a^2+b^2\ge 2ab.$$ That is, $$c^3\ge 2abc.(1)$$ Next, we will show that $a^3+b^3\ge \sqrt{2}abc$. Consider $a^3+b^3=(a+b)(a^2+b^2)-ab(a+b)$ and $$\begin{array}{rl}(a+b)(a^2+b^2)& =(a+b)c^2=(a+b)\sqrt{a^2+b^2}c\\ & \ge 2\sqrt{ab}\sqrt{2ab}c\\ & =2\sqrt{2abc}.\end{array}(2)$$ Now, by the Cauchy-Schwarz inequality, $$\begin{array}{rl}ab(a+b)& \le ab\sqrt{2}\sqrt{a^2+b^2}\\ & =\sqrt{2abc},\end{array}\text{\hspace{1em}(3)}$$ and so $$a^3+b^3\ge 2\sqrt{2abc}-\sqrt{2abc}=\sqrt{2abc}(4)$$ We conclude from (1) and (4) that $$\frac{a^3+b^3+c^3}{abc}\ge 2+\sqrt{2}.(5)$$ This will be an equality when $a=b$.