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let $m,n$ is give positive numbers,and such $$\dfrac{m}{a}+\dfrac{n}{b}=1$$ show that $$a+b+\sqrt{a^2+b^2}\ge 2(m+n+\sqrt{2mn})$$

use this methods:How to find the minimum of $a+b+\sqrt{a^2+b^2}$

let $$a=\dfrac{m(x+y)}{x}.b=\dfrac{n(x+y)}{y}$$ then we have $$a+b+\sqrt{a^2+b^2}- 2(m+n+\sqrt{2mn})=\cdots\ge 0?$$

so I think this is have AM-GM and Cauchy-Schwarz inequality solve it，Thank you

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let $$a=\dfrac{m(x+y)}{x},b=\dfrac{n(x+y)}{y}$$ then $$\Longleftrightarrow \dfrac{m(x+y)}{x}+\dfrac{n(x+y)}{y}+(x+y)\sqrt{\dfrac{m^2}{x^2}+\dfrac{n^2}{y^2}}\ge 2(m+n+\sqrt{2mn})$$ $$\Longleftrightarrow (x+y)(my+nx+\sqrt{(my)^2+(nx)^2})\ge 2xy(m+n+\sqrt{2mn})$$ $$\Longleftrightarrow(x+y)\sqrt{m^2y^2+n^2x^2}\ge (x-y)(my-nx)+2xy\sqrt{2mn}$$ note $$(x+y)^2(m^2y^2+n^2x^2)-((x-y)(my-nx)+2xy\sqrt{2mn})^2=xy(2my-2nx-\sqrt{2mn}(x-y))^2\ge 0$$

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