Finding the minimum of a function of two variables

Question

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??

Answer 1

Hint: Consider vectors $X, A, B, Y$ where $|X| = 7$, $|A| = a$, $|B| = b$, $|Y| = \sqrt{50}$ and $\angle XOA = 45^\circ$, $\angle AOB = 45^\circ$, $\angle BOY = 45^\circ $.

How does your expression relate to these vectors? Think of distances. Apply Cosine rule.

Hence, the minimum is $ \sqrt{7^2 + (\sqrt{50})^2 - 2 \times 7 \times \sqrt{50} \times \cos 135^\circ} = \sqrt{49 + 50 + 70} = 13 $.

Answer 2

Comment to the earlier hint: Consider a triangle XOY where $|OX|=7$ and $|OY|=\sqrt{50}$. Then that expression is equal to $|XY|$. The minimum for $|XY|$ is achieved when $∠XOY=135^∘$. Then the law of cosine give you the minimum for $|XY|$ namely the minimum value for the original expression.