Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$ 
Find maximum value of $x$. $x$ and $y$ are reals.
 A: Try this method of completing the square.
Let $a=\sqrt x$ and $b=\sqrt y$ so that $a^2+b^2=a+b$ and $4a^2-4a+1+4b^2-4b+1=2$ ie $(2a-1)^2+(2b-1)^2=2$
A: Obviously $x,y\ge0$. We can therefore write $x=r^2\cos^2\theta$ and $y=r^2\sin^2\theta$. Then the equation becomes $$r^2(\cos^2\theta+\sin^2\theta)=r(\cos\theta+\sin\theta)$$
with $\theta \in [0,\frac\pi2]$ and $r\gt 0$. Dividing by $r$ and multiplying by $\cos\theta$ we get $$\sqrt x=(\cos\theta+\sin\theta)\cos\theta=\cos^2\theta+\sin\theta\cos\theta$$
Now find the maximum of the expression on the right. Use formula for the sine and cosine for double angle.
A: 
If you set $f(x)=x-\sqrt{x}$ then you want to solve $f(x)=-f(y)$
The maximum $x$ is reached for the line in green intersecting both curves $f$ and $-f$ at the maximal possible altitude.
So we have to solve $f'(y)=0\iff1-\dfrac1{2\sqrt{y}}=0\iff y=\frac 14$
The corresponding $x$ verifies $f(x)=-f(y)=\dfrac 14$ and this is just a quadratic polynomial to solve to find the $x$ value.
The maximum is then $x=\frac{3+2\sqrt{2}}4\approx 1.4571$
A: 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:
$$\color{red}x=\left(\frac{1+\sqrt{3-2\sqrt{2}}}{1+(3-2\sqrt{2})}\right)^2=\frac14\left(1+\sqrt{2}\right)^2\approx 1.457.$$
A: Observe that
$$x+y=\sqrt x+\sqrt y\iff \sqrt x\left(\sqrt x-1\right)=\sqrt y\left(1-\sqrt y\right)$$
Now, for any values $\;x,y>1\;$ , the left side above will be positive whereas the right side will be negative (can you see why?), thus we must restrict the possible natural values...Can you take it from here?
A: Let $f(x)=x-\sqrt x$. The question becomes trying to find two real values $x,y$ so that $f(x)=-f(y)$, then maximising one of these values without regard for what happens to the other. After a bit of thinking analysing derivatives, it becomes clear that we want to minimise $f(x)$ so that $f(y)$, and hence $y$, can be maximised. So just set $f'(x)=0$ to find the local minimum value to be $-1/4$, and hence the maximum value of $y$ satisfies $f(y)=1/4$. This is a simple quadratic equation, solve, and you're done!
A: There are some caveats (we need to assume the existence of a maximum for example) but Lagrange's multipliers method work well here to find the value of the maximum:
Let $f(x,y)=x$ and $g(x,y)=x-\sqrt{x}+y-\sqrt{y}$. We want to find the maximum value of $f$ under constraint $g(x,y)=0$.
We have $\nabla f=\langle 1,0 \rangle$ and $\nabla g = \langle 1-\frac{1}{2\sqrt{x}},1-\frac{1}{2\sqrt{y}} \rangle$. At a maximum, we have $\nabla g=\lambda \nabla f$ for some real number $\lambda$. This implies that $1-\frac{1}{2\sqrt{y}}=0$, that is $y=\frac{1}{4}$.
Using the constraint, we have $x-\sqrt{x}-\frac{1}{4}=0$ which is equivalent to $a^2-a-\frac{1}{4}$ and $\sqrt{x}=a\geqslant 0$. We obtain $a=\frac{1+\sqrt{2}}{2}$, so $x=\frac{(1+\sqrt{2})^2}{4}=\frac{3+2\sqrt{2}}{4}$.

(sloppy) proof of the existence of the global maximum value: The constraint $x+y=\sqrt{x}+\sqrt{y}$ implies that $x\geqslant 0$ and $y\geqslant 0$. Moreover, if $x$ or $y$ is large enough (x>R or y>R for some $R$ that I am too lazy to determine explicitly :p), $x+y>\sqrt{x}+\sqrt{y}$, so we can assume that $x,y \in [0,R]$.
Now, the function $f(x,y)=x$ is continuous and the set $S=\left\{ (x,y)\in [0,R]^2\; \middle|\;  g(x,y)=0\; \right\}$ is compact (closed and bounded in ${\mathbb R}^2$) so, by the Extreme Value Theorem, $f$ attains its global extreme values on $S$. The global minimum value is obviously obtained at $(0,0)$, while the other one is located by Lagrange's method.
