Let $a,b> 0$ and $\frac{8}{a}+\frac{1}{b} =1.$ Find the minimum of $\sqrt{a^2+b^2}.$

(Nanjing University 2021)

Solution: By C-S,

$$2a+b=(2a+b)\left(\frac{8}{a}+\frac{1}{b}\right)\geq 25$$$$\sqrt{a^2+b^2}=\frac{\sqrt{(4+1)(a^2+b^2)}}{\sqrt 5}\geq\frac{2a+b}{\sqrt 5}\geq 5\sqrt 5.$$Equality holds when $a=2b=10.$