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> Find the smallest value of $\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}\quad \quad$ for $a,b$ real and positive.
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What I've done so far:
Let $F(a,b)=\displaystyle \sqrt{49+a^2-7{\sqrt{2}}\ a}+\sqrt{a^2+b^2-{\sqrt{2}}\ ab}+\sqrt{50+b^2-10b}$

Then
$$\begin{align}
\displaystyle F\_a=\frac{2a-7{\sqrt{2}}\ a}{2\sqrt{49+a^2-7{\sqrt{2}}\ a}}+\frac{2a-\sqrt{2}\ b}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}=0\\
\displaystyle F\_b=\frac{2b-{\sqrt{2}}\ a}{2\sqrt{a^2+b^2-{\sqrt{2}}\ ab}}+\frac{2b-10}{2\sqrt{50+b^2-10b}}=0
\end{align}$$ 
I then tried to simplify the two equations as follows:
$$\begin{align}
\left({2a-7{\sqrt{2}}\ a}\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)+\left({2a-\sqrt{2}\ b}\right)\left(\sqrt{49+a^2-7{\sqrt{2}}\ a}\right)=0\\
\left({2b-{\sqrt{2}}\ a}\right)\left(\sqrt{50+b^2-10b}\right)+\left(2b-10\right)\left(\sqrt{a^2+b^2-{\sqrt{2}}\ ab}\right)=0\\
\end{align}$$
But then I'm stuck.

*Do I need to include a constraint?(i.e $a$ and $b$ are positive and real); if so what will the constraint equation be?*

*Is there an altogether different way to do this other than using Lagrange??*