<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] Now that someone has worked out the place where the minimum is achieved ($a:b = 4:3$), one can purposely use Cauchy-Schwarz/AM-GM.
$$(a^2+b^2)(16+9) \ge (4a+3b)^2 \Rightarrow \sqrt{a^2+b^2} \ge \frac{4a+3b}{5}$$
Then
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5}$$
Again Cauchy-Schwarz,
$$(9a+8b)(\frac{4}{a} + \frac{2}{b}) \ge (6+4)^2 = 100 \Rightarrow 9a+8b \ge 50$$
Therefore
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5} \ge 10$$
</s>