Minimum value of an expression in two variables

Question

$$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}$$

$x$ and $y$ are positive real numbers, what is the minimum value?

I have tried finding the minimum value of the first expression and the third expression, with the first expression being $\frac{7}{\sqrt{2}}$ while the second expression being $5$, I do not know how to move on from this (note that the answer is an integer from $0$ to $999$).

Answer 1

let $$f(x,y)=\sqrt{x^2-7\sqrt{2}x+49}+\sqrt{x^2+y^2-\sqrt{2}xy}+\sqrt{50+y^2-10y}$$ then solve $$f_x=1/2\,{\frac {2\,x-7\,\sqrt {2}}{\sqrt {{x}^{2}-7\,\sqrt {2}x+49}}}+1/2 \,{\frac {2\,x-\sqrt {2}y}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}} =0$$ and $$f_y=1/2\,{\frac {2\,y-\sqrt {2}x}{\sqrt {{x}^{2}+{y}^{2}-\sqrt {2}xy}}}+1/ 2\,{\frac {2\,y-10}{\sqrt {{y}^{2}-10\,y+50}}} =0$$

Answer 2

Let $OABCD$ be a pentagon such that $OA=5\sqrt2$, $OB=y$, $OC=x$, $OD=7$, $\measuredangle AOB=45^{\circ}$, $\measuredangle BOC=45^{\circ}$ and $\measuredangle COD=45^{\circ}$.

Thus, $$\sqrt{x^2-7\sqrt{2}x+49} + \sqrt{x^2+y^2-\sqrt{2}xy} + \sqrt{50+y^2-10y}=$$ $$=DC+CB+BA\geq AD=\sqrt{(5\sqrt2)^2+7^2-2\cdot5\sqrt2\cdot7\cos135^{\circ}}=13.$$ The equality occurs, when $\{B,C\}\subset AD,$ which says that $13$ is a minimal value.

Done!