Team Selection Test for IMO 2023

The answer is 6. The equality holds when $$(a,b,c,d)=\left(\sqrt[4]{6},\sqrt[4]{\frac{27}{2}},\frac{\sqrt{6}}{2},1\right).$$ Solution 1. By Cauchy-Schwarz inequality we get $$\left[(a^2+3d^2)+(b^2+2c^2)\right]\left[\frac{a^2}{a^2+3d^2}+\frac{b^2}{b^2+2c^2}\right]\ge (a+b{)}^2$$ and $$\left[(2a^2+6d^2)+(3b^2+6c^2)\right]\left[\frac{3d^2}{a^2+3d^2}+\frac{2c^2}{b^2+2c^2}\right]\ge 6(c+d{)}^2$$ Then it follows that $$\begin{gathered} S:=\frac{(a+b{)}^2}{a^2+b^2+2c^2+3d^2}+\frac{6(c+d{)}^2}{2a^2+3b^2+6c^2+6d^2} \\ \le \frac{a^2}{a^2+3d^2}+\frac{b^2}{b^2+2c^2}+\frac{3d^2}{a^2+3d^2}+\frac{2c^2}{b^2+2c^2}=2 \end{gathered}$$ Finally by Arithmetic-Geometric Mean inequality, we have $$\frac{6(a+b{)}^2(c+d{)}^2}{(a^2+b^2+2c^2+3d^2)(2a^2+3b^2+6c^2+6d^2)}\le \frac{S^2}{4}\le 1$$ and hence we are done.

Solution 2. It follows from Cauchy-Schwarz inequality that $$(a^2+b^2+2c^2+3d^2)(6c^2+6d^2+3b^2+2a^2)$$ $$\ge 6(ad+bc+ac+bd{)}^2=6(a+b{)}^2(c+d{)}^2$$ and hence the minimal value of the expression is 6.