Find the maximum value of $a+b$ The question:

Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
  $$a+\sqrt{b} = b + \sqrt{a}$$

Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
 A: We have
$$a+\sqrt b=b+\sqrt a\iff a-b=(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)=\sqrt a-\sqrt b.$$
Then $a=b$ (which is not allowed) or $\sqrt a+\sqrt b=1$.
Now,
$$a+b=a+(1-\sqrt a)^2$$ has an extremum found by taking the derivative on $a$,
$$1-\frac{1-\sqrt a}{\sqrt a}=0$$
which gives $a=\dfrac14$. But it turns out that this is a minmum.
Then we also have to try the values at the boundaries of the domain: $a=0,b=1$ and $a=1,b=0$ both yield
$$a+b=1$$
which is the searched maximum.
A: Observe that
$$a-\sqrt a=b-\sqrt b\implies a-b=\sqrt a-\sqrt b\iff\frac{(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)}{\sqrt a-\sqrt b}=1\iff$$
$$\iff \sqrt a+\sqrt b=1\;(\text{ assuming $\,a\neq b\,$)}\implies b=(1-\sqrt a)^2$$
So you need the maximum of $\;f(a):=a+b=a+(1-\sqrt a)^2=2a-2\sqrt a+1\;$ ...can you take it from here?
A: As
$$
a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=\sqrt{a}-\sqrt{b}
$$
then with $a\ne b$ it gives $\sqrt{a}+\sqrt{b}=1$. Now you have to maximize 
$$
(\sqrt{a})^2+(\sqrt{b})^2
$$
subject to $\sqrt{a}+\sqrt{b}=1$.
A: For $a\ne b,$
$$\sqrt a+\sqrt b=1$$
WLOG $a=\cos^4t,b=\sin^4t$
$$a+b=1-2\sin^2t\cos^2t=1-\dfrac{\sin^22t}2$$
Now $0\le\sin^22t\le1$
Or $a+b=1-\dfrac{1-\cos4t}4=\dfrac{3+\cos4t}4$
Now $-1\le\cos4t\le1$
A: WLOG, let $b=ax,x\ge 0$. Then:
$$a+\sqrt{b} = b + \sqrt{a} \iff a+\sqrt{ax}=ax+\sqrt{a} \iff \\
\sqrt{a}(\sqrt{x}-1)=a(x-1) \stackrel{x\ne 1}{\iff} 1=\sqrt{a}(\sqrt{x}+1) \iff a=\frac{1}{(\sqrt{x}+1)^2}.$$
Hence:
$$a+b=a+ax=\frac{1}{(\sqrt{x}+1)^2}+\frac{x}{(\sqrt{x}+1)^2}=\frac{x+1}{(\sqrt{x}+1)^2}\le 1, x\ge 0.$$
Equality occurs for $x=0$.
A: Calling 
$$
a = x^2\\
b = y^2
$$
the problem reads now
$$
\max (x^2+y^2)\;\;\mbox{s.t.}\;\; x^2-y^2=x-y\Rightarrow x+y=1
$$
now the problem is reduced to: 
Find $r$ in $C\to x^2+y^2-r^2=0$ such that $C$ is tangent  to $x+y = 1$
A: A graph in this situation greatly helps.
Graphing the squared equation $(a-\sqrt a)^2=(b-\sqrt b)^2$ in the $aOb$ Cartesian plane gives a bean-like shape, touching the axes at $(1,0)$ and $(0,1)$. It can be verified that the upper-right part of the graph is an artifact of the squaring. Therefore the maximum point should occur at $(1,0)$ and $(0,1)$, since you can calculate the slope of the tangent line there to be zero and infinite, respectively.
However this is not a complete and rigorous solution. To complete it, you need to prove that $0\le a,b\le1$, which you've done so; and then you can demonstrate that the curve is concave upwards (when using the coordinate system $uOv$, where $u,v=a\pm b$), by calculating the implicit derivative. Finally you can conclude that the maximum value occurs at the endpoints.
