Mongolian National Mathematical Olympiad

By the Cauchy-Schwarz inequality, we have $$3\sqrt{1+5a^2}=\sqrt{2^2+(\sqrt{5}{)}^2}\cdot \sqrt{1+5a^2}\ge 2+5a$$ and $$3\sqrt{2+b^2}=\sqrt{(2\sqrt{2}{)}^2+1}\cdot \sqrt{2+b^2}\ge 4+b.$$ Hence $$\sqrt{1+5a^2}+5\sqrt{2+b^2}\ge \frac{2+5a}{3}+\frac{5(4+b)}{3}=9.$$ Equality holds for $a=b=1/2$.