Let $x + \sqrt{x^2+1}$ take on the value $\alpha$ (note that $\alpha > 0$ for all $x$) and solve the given equation for $y$: \[\alpha (y+\sqrt{y^2+4})=7 \Rightarrow y = \frac{7}{2\alpha }-\frac{2\alpha }{7}\]In order to facilitate the algebra let $\beta = \frac{7}{2\alpha } > 0$, for all $x$.We are looking for the value of the sum: $x\sqrt{y^2+4}+y\sqrt{x^2+1}$. Elaborating on each term:(i). \[x\sqrt{y^2+4} = x\sqrt{\left ( \beta -\frac{1}{\beta } \right )^2+4}=x\sqrt{\left ( \beta +\frac{1}{\beta } \right )^2}= x\left ( \beta +\frac{1}{\beta } \right )\]
(ii). \[y\sqrt{x^2+1} = y(\alpha -x)=\left (\beta -\frac{1}{\beta } \right )\left ( \alpha -x \right )\]Summing the terms:
Here is a little puzzle from the book 100 Geometric Games by
Pierre Berloquin.
The side of a small square is one meter long and the side of
a larger square one and a half meters long. One vertex of the
large square is at the center of the small square. The side of
the large square cuts two sides of the small square into one-
third parts and two-thirds parts.
What is the area where the squares overlap?