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?

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.
