IF $f(0)=0$, then with $x:=0$, we get \[f\big(f(y)\big)=f(y)\,.\;\;\;\text{(1)}\]With $y:=x$, we now have \[f\left(\frac{3x+f(x)}{2}\right)=2x+f(x)-f\big(f(x)\big)=2x\,.\]Consequently, $f$ is onto, and therefore, (1) implies that $f(x)=x$ (which can be readily checked to be a solution).

[mod edit: also posted here.]

I think we can say $f(0)=0$ in an easier way . by putting $x=f(x)$ and $y=x$ we get $f(\frac{f(f(x))+f(x)}{2}+x)=2x$ and so $f$ is surjective . this implies that there is $a$ that $f(a)=0$ . now put $x=a$ and $y=\frac{a}{2}$ . we get $f(f(\frac{a}{2}))=-2a$ . since $a$ and $-2a$ are both non-negative , we get $a=0$ and $f(0)=0$ .

goodar2006 wrote: by putting $x=f(x)$ and $y=x$ we get $f(\frac{f(f(x))+f(x)}{2}+x)=2x$ and so $f$ is surjective . i think it is wrong can you write it more clearly?

yes , you are right . we get $f(\frac{f(f(x))+f(x)}{2}+x)=2f(x)$ . thanks for noticing me .

We can show that $f(0)=0$ in this way: because $f$ is non-decreasing we have $f(0) \le f( \frac{f(0)}{2})=f(f(0))-f(0)$ so $f(f(0)) \ge 2f(0)$. By putting $x=f(0), y=0$ we get $f( \frac{f(0)+f(f(0))}{2})=2f(0) \le f(f(0))$. That means $f(0 )\ge \frac{f(0)+f(f(0))}{2}$ or $f( \frac{f(0)+f(f(0))}{2})=2f(0)=f(f(0))$. If first then obviously $f(0)=0$. Let second be correct. By placing $x=0, y= \frac{f(0)}{2}$ we obtain $f(f(0))=f(f( \frac{f(0)}{2}))-f(0);$ $f(f(0))=f(f(f(0))-f(0))-f(0);$ $f(f(0))=f(f(0))-f(0);$ $ f(0)=0$

Let $P(x,y)$ denote the assertion that $f\left(\frac{x+f(x)}{2}+y\right)=2x-f(x)+f(f(y))$. First suppose to the contrary that some $a<b$ exist such that $\ell=f(a)=f(b)$. Then $f(x)=\ell$ for all $a\le x\le b$, so for sufficiently small $\epsilon$, \[P(a,b),P(a+2\epsilon,b-\epsilon)\implies 2a-\ell+f(\ell)=f\left(\frac{a+\ell}{2}+b\right)=2(a+2\epsilon)-\ell+f(\ell),\]a contradiction. Thus $f$ is injective, so $a<b\implies f(a)<f(b)$ and $f(a)\ge f(b)\implies a\ge b$, whence \[P(0,r)\implies f(f(r))=f(0)+f\left(\frac{f(0)}{2}+r\right)\implies f(r)\ge r+\frac{f(0)}{2}\forall{r}.\]Now assume for the sake of contradiction that $c=f(0)/2>0$. Then \begin{align*} P(f(r),r)\implies 2f(r) &= f\left(\frac{f(r)+f(f(r))}{2}+r\right) \\ &\ge \frac{r+c+r+2c}{2}+r+c \\ &= 2r+\frac{5c}{2}\implies f(r)\ge r+\frac{5c}{4} \end{align*}for all $r$, which is clearly impossible (since $(5/4)^n$ tends to infinity). Hence $f(0)=0$, so \[P(0,r)\implies f(f(r))=f(r)\implies f(r)=r\forall r,\]as desired.

My solution is different from the upper ones. First, like # 6 , we can get f(0)=0 then, put x=0 , f(y)=f(f(y)), thus f( (x+f(x)) / 2 ) + y = 2x - f(x) +f(y) replacing y by f(y), we have f( x+f(x) / 2 ) = 2x - f(x) we can have the following discusion. (1) if x >= f(x) x>=( x+f(x) )/2 2x= f(x)+f( (x+f(x)) / 2 ) <=2f(x) thus we have f(x)=x (2) if x< f(x) like (1) we'll find out that this condition is wrong. by the upper discusion , the answer is f(x)=x

Doesn't look like we even need monotonicity Let $c=f(f(0))$ and put $y=0$ in the given equation to get$$f\left(\frac{x+f(x)}2\right)=2x-f(x)+c.\qquad (\star)$$ Claim 1. $f(x)\le x+c$ for all $x\ge 0$.

Claim 2. $f(x)=x+\tfrac25c$ for all $x\ge 0$.

Now using $f(x)=x+\tfrac25c$ in the given equation immediately gives $c=0$, so $\boxed{f(x)\equiv x}$ which is indeed a solution. $\blacksquare$

I have a quicker solution I believe, and also more beautiful IMO because it only proves the existance of a fixed point and not that $f(0)=0$. First, lets prove that $f(x)\geq x$ for all $x$. Let us assume by contradiction that $f(a)<a$. substituting $(a,0)$ into the equation, we get $f(\frac{a+f(a)}{2})+f(a)=2a+f(f(0))$. Yet, since $\frac{a+f(a)}{2}<a$, LHS =RHS $\leq 2f(a)<2a$. Thus, we get $f(f(0))<0$ which is a contradiction. Now, let us prove that a fixed point exists. By contradiction, assume that a fixed point doesn't exist. Thus, let the minimum value of $f(x)-x$ be $c>0$. Thus, $f(c)\geq x+c$, for which equality is attainable. We get $2x-f(x)+f(f(y))=f(\frac{x+f(x)}{2}+y)\geq \frac{x+f(x)}{2} +c \geq x+y+1.5c$. Thus, $f(f(y))\geq y+1.5c +f(x)-x \geq y+2.5c$ so $f(f(y))\geq y+2.5c$ for all values of $y$. Now, plug in $(f(y),y)$ to get $2f(y)=f(\frac{f(y)+f(f(y))}{2}+y)\geq \frac{f(y)+f(f(y))}{2}+c+y \geq \frac{f(y)}{2}+\frac{y+2.5c}{2}+c+y$. From here, we obtain that $f(y)\geq y+1.5c$ which is a contradiction because $c>0$ and there must be an equality case of $f(y)=y+c$. Now, assume the fixed point is $r$, so $r=f(r)=f(f(r))$. Trying $(f(x),r)$ gets you $2x-f(x)+r=f(\frac{x+f(x)}{2}+r)\geq \frac{x+f(x)}{2}+r$. Thus $f(x)\leq x$ for all $x$ so $f(x)=x$ at all points $x$ and we're done!

MessingWithMath wrote: By contradiction, assume that a fixed point doesn't exist. Thus, let the minimum value of $f(x)-x$ be $c>0$. It's possible that $\{f(x)-x | x\ge 0\}$ has no minimum (e.g. when $f(x)\equiv x+e^{-x}.$)

The answer is $f(x) \equiv x$, which clearly works. Let $P(x,y)$ be the given assertion. Claim: $f(x) \ge x ~~ \forall ~ x \in \mathbb R_{\ge 0}$. Proof: FTSOC, $x > f(x)$ for some $x$. Put $y = \frac{x-f(x)}{2}$ in $P(x,y)$. Using $f(f(y)) \ge 0$ we get \begin{align*} 2x - f(x) \le f \left( \frac{x+f(x)}{2} + \frac{x-f(x)}{2} \right) = f(x) \implies x \le f(x) \end{align*}So we obtain a contradiction. $\square$ $P(0,y)$ gives $$ f \left( \frac{f(0)}{2} + y \right) = f(f(y)) - f(0) $$Letting $f(0) = 2c$ we get $$ f(c+y) + 2c = f(f(y)) \qquad \qquad (1) $$Let functions $$ g(x) = \frac{x+f(x)}{2} ~~,~~ h(x) = 2x - f(x) $$Note $g$ is strictly increasing (as $f$ is monotone). Fix any $a > b \ge 0$. Let $$ r = g(a) - g(b) ~~,~~ s = h(a) - h(b) $$Consider any $y_1,y_2$ such that $y_2 - y_1 = r$. Comparing $P(a,y_1)$ and $P(b,y_2)$ gives $$ h(a) + f(f(y_1)) = h(b) + f(f(y_2)) \implies f(f(y_2)) - f(f(y_1)) = s $$Invoking $(1)$ gives $$ f(y+r) = f(y) + s ~~ \forall ~ y \ge c \qquad \qquad (2)$$Now using $f(x) \ge x$ we easily get $s \ge r$. Basically, for any pair $(m,n)$ with $m > n \ge 0$ we obtain \begin{align*} g(m) - g(n) \le h(m) - h(n) \\ \implies g(m) - h(m) \le g(n) - h(n) \\ \implies \frac{3}{2} \left( f(m) - m \right) \le \frac{3}{2} \left(f(n) - n \right) \\ \implies f(m) - f(n) \le m -n \end{align*}For a large $n$, putting $m = n+r$ and using $(2)$ gives $$ s \le r $$Combining this with $r \le s$ forces $r=s$. So for any $a > b \ge 0$ we have $$ g(a) - g(b) = h(a) - h(b) $$As before, this forces $$ f(a) - f(b) = a-b $$We conclude $f(x) \equiv x+d$ for some constant $d$. From $(1)$, it isn't hard to conclude $d=0$. Hence $f(x) \equiv x$. $\blacksquare$