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
 A: 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
$$a+b+\sqrt{a^2+b^2} \ge \frac{9a+8b}{5} \ge 10$$
A: 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
$$|OC|+|OD|+|CD|\ge|OE|+|OF|+|EF|=|OA|+|OB|+AB|=10$$
A: let
$$a=\dfrac{2(x+y)}{x},b=\dfrac{x+y}{y},x,y>0$$
then
$$C=a+b+\sqrt{a^2+b^2}=\dfrac{(x+y)(x+2y+\sqrt{4y^2+x^2})}{xy}$$
so
$$C-10=\dfrac{x^2-7xy+2y^2+(x+y)\sqrt{x^2+4y^2}}{xy}$$
and note
$$(x+y)^2(x^2+4y^2)-(x^2-7xy+2y^2)^2=4xy(2x-3y)^2\ge 0$$
so
$$a+b+\sqrt{a^2+b^2}\ge 10$$
A: 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.
A: 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 $$a+b+\sqrt{a^2+b^2}= \frac{10}{3}+\frac{5}{2}+\sqrt{\left(\frac{10}{3}\right)^2+\left(\frac{5}{2}\right)^2} = 10\textrm.$$
