How to find the minimum of $a+b+\sqrt{a^2+b^2}$ let $a,b>0$, and such
$$\dfrac{2}{a}+\dfrac{1}{b}=1$$
Find this minimum
$$a+b+\sqrt{a^2+b^2}$$
My try: since
$$2b+a=ab$$
so
$$a+b+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+2ab}+\sqrt{a^2+b^2}=\sqrt{a^2+b^2+4b+2a}+\sqrt{a^2+b^2}$$
then I can't
maybe this problem can use AM-GM or Cauchy-Schwarz inequality solve it.Thank you very much

let
$$a=\dfrac{2(x+y)}{x},b=\dfrac{x+y}{y},x,y>0$$
then
$$C=a+b+\sqrt{a^2+b^2}=\dfrac{(x+y)(x+2y+\sqrt{4y^2+x^2})}{xy}$$
so
$$C-10=\dfrac{x^2-7xy+2y^2+(x+y)\sqrt{x^2+4y^2}}{xy}$$
and note
$$(x+y)^2(x^2+4y^2)-(x^2-7xy+2y^2)^2=4xy(2x-3y)^2\ge 0$$
so
$$a+b+\sqrt{a^2+b^2}\ge 10$$
