Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$ 
Find maximum value of $x$. $x$ and $y$ are reals.

Since $x+ y = \sqrt{x} + \sqrt{y}$ is symmetric, WLOG, we can consider $x\ge y\ge 0$ or $y=ax, 0\le a\le 1$. Then:
$$x+ ax = \sqrt{x} + \sqrt{ax} \Rightarrow \color{red}{\sqrt{x}}=\frac{1+\sqrt{a}}{1+a} \to \text{max s.t. $0\le a\le 1$}\\
f'(a)=\left(\frac{1+\sqrt{a}}{1+a}\right)'=0 \Rightarrow \frac{\frac1{2\sqrt{a}}(1+a)-(1+\sqrt{a})}{(1+a)^2}=0 \Rightarrow \\
a+2\sqrt{a}-1=0 \Rightarrow a=3-2\sqrt{2}\approx 0.17.$$
Note that the function $f(a)$ is continuous at $0\le a\le 1$ and $f(0)=f(1)=1$ and $f(3-2\sqrt{2})=\frac12(1+\sqrt{2})\approx 1.207$. Hence it must be the maximum point (see Desmos graph). 
So, the maximum value of $x$ is:
$$\color{red}x=\left(\frac{1+\sqrt{3-2\sqrt{2}}}{1+(3-2\sqrt{2})}\right)^2=\frac14\left(1+\sqrt{2}\right)^2\approx 1.457.$$
