Solution of a function equation $f(x) + f(y) = f(x + y + 2f(xy))$ Find all functions $f:\mathbb{R}_{\geq0}\rightarrow \mathbb{R}_{\geq0}$ which satisfies that for $x,y\in\mathbb{R}_{\geq0}$, $$f(x)+f(y)=f(x+y+2f(xy))$$
I spent quite some time trying to solve it but didn't succeed.  It is clear that it has a solution $f(x) = 0$ and $f(x) = \sqrt{x}$ but the problem is whether they are all.  Interesting related problem is that if we restrict $f$ to be continuous, or analytic, what will happen.
I tried to : Using Hamel basis to show the solutions are infinite.  Failed. Also I tried to reduce it to Cauchy's 1-4 equations but didn't succeed.  In the corse of it, I found interesting works of Aczel, Erdos and even Putnum, but they are not directly related, I guess.
Any idea?  By the way, this is not a homework.  A math fun came up with the problem but couldn't solve.

1. Main Claim
We write $\mathsf{Sol}$ for the set of all solutions of the functional equation.

Claim. If $f \in \mathsf{Sol}$ is injective, then $f(x) = \sqrt{x}$.

Proof. Assume that $f \in \mathsf{Sol}$ is injective. Applying the functional equation in two ways, we find that
\begin{align*}
f(x) + f(y) + f(1)
&= f(x) + f(y + 1 + 2f(y)) \\
&= f(x + y + 1 + 2f(y) + \boxed{ 2f(xy + x + 2xf(y)) })
\end{align*}
and
\begin{align*}
f(x) + f(y) + f(1)
&= f(x + y + 2f(xy)) + f(1) \\
&= f(x + y + 2f(xy) + 1 + 2f(x + y + 2f(xy))) \\
&= f(x + y + 1 + \boxed{ 2f(xy) + 2f(x) } + 2f(y)).
\end{align*}
By the injectivity assumption, we have
\begin{align*}
f(xy + x + 2xf(y))
&= f(xy) + f(x) \\
&= f(xy + x + 2f(x^2y)).
\end{align*}
Stripping $f$ off both sides of the identity above, we find that
$$ f(x^2y) = xf(y). $$
So it follows that $f(x) = f(1)\sqrt{x}$, and plugging this back to the functional equation shows that $f(1) = 1$. Therefore $f(x) = \sqrt{x}$. ////

2. Some extra observations

Lemma 1. Let $f \in \mathsf{Sol}$. Then
  
  
*
  
*$f(x) + nf(1) = f(x + 2nf(x) + n + n(n-1)f(1))$ for all $x\geq0$ and $n \geq 0$.
  
*$f(0) = 0$.
  

Proof. (1) Let us write $x_n = x + 2nf(x) + n + n(n-1)f(1)$. Since $x_0 = x$, the $n = 0$ case is trivial. Now assume that $f(x) + nf(1) = f(x_n)$. Then
\begin{align*}
f(x) + (n+1)f(1)
&= f(x_n) + f(1) \\
&= f(x_n + 1 + 2f(x_n)) \\
&= f(x_n + 1 + 2f(x) + 2nf(1))
 = f(x_{n+1})
\end{align*}
and hence the claim follows from the mathematical induction.
(2) From the previous part with $x = 0$ and $n = 2$, it follows that
$$ f(0) + 2f(1) = f(4f(0) + 2 + 2f(1)). $$
Now using the identity $f(x + 2f(0)) = f(0) + f(x)$ twice, we find that
$$ f(4f(0) + 2 + 2f(1)) = 2f(0) + f(2 + 2f(1)) = 2f(0) + 2f(1). $$
Comparing two formulas yields $f(0) = 0$. ////

Corollary 2. If $f \in \mathsf{Sol}$ is non-decreasing, then 
  
  
*
  
*$\lim_{x\downarrow 0} f(x) = 0$.
  
*Either $f\equiv 0$ or $f(x) = \sqrt{x}$.
  

Proof. (1) Write $\ell = \lim_{x\downarrow 0}f(x)$. We claim that $\ell = 0$. Indeed, let $\epsilon > 0$ and write
\begin{align*}
\epsilon_1 &= 2f((2+2f(1))\epsilon) - 2\ell + \epsilon, \\
\epsilon_2 &= 2f((2 + 2f(1) + 2\ell + \epsilon_1)\epsilon) - 2\ell + \epsilon_1 + \epsilon.
\end{align*}
Then mimicking the previous proof, we have
\begin{align*}
2\ell + 2f(1)
&\leq 2f(\epsilon) + f(2+2f(1)) \\
&= f(\epsilon) + f(2 + 2f(1) + 2\ell + \epsilon_1 ) \\
&= f(2 + 2f(1) + 4\ell + \epsilon_2) \\
&\leq f(2 + 2f(1) + 4f(\epsilon_2) + \epsilon_2)
 = f(\epsilon_2) + 2f(1).
\end{align*}
Here, intermediate equalities follow from the functional equation together with the definition of $\epsilon_1$ and $\epsilon_2$, and the last equality follows from Lemma 1.(1). Since $\epsilon_2 \downarrow 0$ as $\epsilon \downarrow 0$, taking $\epsilon \downarrow$ to the above bound tells that $2\ell + 2f(1) \leq \ell + 2f(1)$ and hence $\ell = 0$.
(2) If $f(a) = 0$ for some $a > 0$, then $f(2a) \leq f(2a+2f(a^2)) = 2f(a) = 0$ and repeating this procedure tells that $f(2^n a) = 0$ for all $n \geq 0$. Since $f$ is non-decreasing, this implies that $f \equiv 0$.
So we may assume that $f(x) > 0$ for all $x > 0$. Then for any $0 \leq x < y$, there exists $t > 0$ such that $y - x > t + 2f(t x)$ by the previous part. Then
$$ f(y) \geq f(x + t + 2f(t x)) = f(x) + f(t) > f(x) $$
and hence $f$ is strictly increasing. Therefore by the main claim, $f(x) = \sqrt{x}$. ////

Corollary 3. If $f \in \mathsf{Sol}$ is continuous, then either $f\equiv 0$ or $f(x) = \sqrt{x}$.

Proof. In view of Corollary 2, it suffices to show that $f$ is non-decreasing. Indeed, fix $x \geq 0$. Since $t \mapsto t + 2f(xt)$ is continuous and takes every value in $\mathbb{R}_{\geq 0}$, for each $y > x$ we can choose $t > 0$ such that $t + 2f(xt) = y - x$. Then
$$ f(y) = f(x + t + 2f(xt)) = f(x) + f(t) \geq f(x). $$
Therefore $f$ is non-decreasing and hence the conclusion follows. ////
