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?
Best Answer
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?