APMO

We first prove that $f(x)\ge 2x$ for all $x>0$. Suppose, for the sake of contradiction, that $f(y)<2y$ for some positive $y$. Choose $x$ such that $f((c+1)x+f(y))$ and $f(x+2y)$ cancel out, that is, $$(c+1)x+f(y)=x+2y⟺x=\frac{2y-f(y)}{c}$$ Notice that $x>0$ because $2y-f(y)>0$. Then $2cx=0$, which is not possible. This contradiction yields $f(y)\ge 2y$ for all $y>0$.

Now suppose, again for the sake of contradiction, that $f(y)>2y$ for some $y>0$. Define the following sequence: $a_0$ is an arbitrary real greater than $2y$, and $f\left(a_n\right)=f\left(a_{n-1}\right)+2cx$, so that $$\{\begin{array}{r}(c+1)x+f(y)=a_n\\ x+2y=a_{n-1}\end{array}⟺x=a_{n-1}-2y\text{ and }{a}_n=(c+1)\left(a_{n-1}-2y\right)+f(y).$$ If $x=a_{n-1}-2y>0$ then $a_n>f(y)>2y$, so inductively all the substitutions make sense.

For the sake of simplicity, let $b_n=a_n-2y$, so $b_n=(c+1)b_{n-1}+f(y)-2y(\ast )$. Notice that $x=b_{n-1}$ in the former equation, so $f\left(a_n\right)=f\left(a_{n-1}\right)+2c{b}_{n-1}$. Telescoping yields $$f\left(a_n\right)=f\left(a_0\right)+2c\sum _{i=0}^{n-1}{b}_i.$$ One can find $b_n$ from the recurrence equation $(\ast ):b_n=\left(b_0+\frac{f(y)-2y}{c}\right)(c+1{)}^n-\frac{f(y)-2y}{c}$, and then $$\begin{array}{rl}f\left(a_n\right)& =f\left(a_0\right)+2c\sum _{i=0}^{n-1}\left(\left(b_0+\frac{f(y)-2y}{c}\right)(c+1{)}^i-\frac{f(y)-2y}{c}\right)\\ & =f\left(a_0\right)+2\left(b_0+\frac{f(y)-2y}{c}\right)\left((c+1{)}^n-1\right)-2n(f(y)-2y).\end{array}$$ Since $f\left(a_n\right)\ge 2a_n=2b_n+4y$, $$\begin{array}{rl}& f\left(a_0\right)+2\left(b_0+\frac{f(y)-2y}{c}\right)\left((c+1{)}^n-1\right)-2n(f(y)-2y)\ge 2b_n+4y\\ =& 2\left(b_0+\frac{f(y)-2y}{c}\right)(c+1{)}^n-2\frac{f(y)-2y}{c},\end{array}$$ which implies $$f\left(a_0\right)+2\frac{f(y)-2y}{c}\ge 2\left(b_0+\frac{f(y)-2y}{c}\right)+2n(f(y)-2y),$$ which is not true for sufficiently large $n$. A contradiction is reached, and thus $f(y)=2y$ for all $y>0$. It is immediate that this function satisfies the functional equation.

After proving that $f(y)\ge 2y$ for all $y>0$, one can define $g(x)=f(x)-2x,g:{\mathbb{R}}_{>0}\to {\mathbb{R}}_{\ge 0}$, and our goal is proving that $g(x)=0$ for all $x>0$. The problem is now rewritten as $$\begin{array}{rl}& g((c+1)x+g(y)+2y)+2((c+1)x+g(y)+2y)=g(x+2y)+2(x+2y)+2cx\\ ⟺& g((c+1)x+g(y)+2y)+2g(y)=g(x+2y)\end{array}\text{\hspace{1em}(1)}$$ This readily implies that $g(x+2y)\ge 2g(y)$, which can be interpreted as $z>2y⟹g(z)\ge 2g(y)$, by plugging $z=x+2y$.

Now we prove by induction that $z>2y⟹g(z)\ge 2m\cdot g(y)$ for any positive integer $2m$. In fact, since $(c+1)x+g(y)+2y>2y,g((c+1)x+g(y)+2y)\ge 2m\cdot g(y)$, and by (??), $$g(x+2y)\ge 2m\cdot g(y)+2g(y)=2(m+1)g(y),$$ and we are done by plugging $z=x+2y$ again.

The problem now is done: if $g(y)>0$ for some $y>0$, choose a fixed $z>2y$ arbitrarily and and integer $m$ such that $m>\frac{g(z)}{2g(y)}$. Then $g(z)<2m\cdot g(y)$, contradiction.