SAUDI ARABIAN IMO Booklet 2023

By the quadratic-arithmetic mean inequality, we have $$1+2t^2\ge \frac{1}{2}(1+t\sqrt{2}{)}^2.$$ It follows that $$\begin{array}{rl}{x}^2\sqrt{1+2y^2}+y^2\sqrt{1+2x^2}& \ge x^2\cdot \frac{1+y\sqrt{2}}{2}+y^2\cdot \frac{1+x\sqrt{2}}{2}\\ & =\frac{x^2+y^2+xy(x+y)\sqrt{2}}{\sqrt{2}}\\ & \ge \frac{2xy+xy(x+y)\sqrt{2}}{\sqrt{2}}=xy(x+y+\sqrt{2}).\end{array}$$ This is what we need to prove. $\square$