<s>[INST] 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
 [/INST] since 
$$\dfrac{1}{a}+\dfrac{2}{b}=1$$
then a straight line
$$\dfrac{x}{a}+\dfrac{y}{b}=1$$ cross $P(1,2)$

then
$$a+b+\sqrt{a^2+b^2}=|OA|+|OB|+|AB|$$
In the follow 

we have
$$|OC|+|OD|+|CD|\ge|OE|+|OF|+|EF|=|OA|+|OB|+AB|=10$$
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