<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] As you stated, $2b + a = ab$. We can get this even nicer: $a = \frac{2b}{b - 1}$.
Substitute into the square root expression:
$$
\begin{align*}
a + b + \sqrt{a^2 + b^2} &= \frac{2b}{b - 1} + b + \sqrt{\left( \frac{2b}{b - 1} \right)^2 + b^2} \\
&= \frac{b^2 + b}{b - 1} + \sqrt{ \frac{b^4 - 2b^3 + 5b^2}{(b - 1)^2} } \\
&= \frac{b^2 + b}{b - 1} + \frac{b}{b - 1} \sqrt{ b^2 - 2b + 5 } \\
&= \frac{b}{b - 1} (b + 1 + \sqrt{ b^2 - 2b + 5 }) \\
\end{align*}
$$
From here I don't see any other way to find the minimum but calculus. Take the derivative and set equal to zero. But it's really messy. On the other hand, Wolfram Alpha gives a minimum of $\frac{5}{2}$ for it.
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