<s>[INST] 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?
 [/INST] 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 $x=0$.
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