<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] Lagrange multipliers seems like a good approach. Solve the system of equations $$\displaystyle \frac{\partial }{\partial a}\left(a+b+\sqrt{a^2+b^2}-\lambda  \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,
$$\frac{\partial }{\partial b}\left(a+b+\sqrt{a^2+b^2}-\lambda  \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm,$$,
$$\frac{\partial }{\partial \lambda }\left(a+b+\sqrt{a^2+b^2}-\lambda  \left(\frac{2}{a}+\frac{1}{b}-1\right)\right)=0\textrm.$$
Mathematica gives the solution as $\left(\lambda = -10, a= \frac{10}{3},b= \frac{5}{2}\right)$, which gives the minimum as $$a+b+\sqrt{a^2+b^2}= \frac{10}{3}+\frac{5}{2}+\sqrt{\left(\frac{10}{3}\right)^2+\left(\frac{5}{2}\right)^2} = 10\textrm.$$
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