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?

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 $\sqrt{a}+\sqrt{b}=1$.
