The answer is $f(x)=2x$ only, which clearly works. Now we prove that this is the only solution. Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for any $y$, then from $P(x,\tfrac{2y-f(y)}{c})$ we find that $2cx=0$: contradiction. Therefore we may let $f(x)=g(x)+2x$, where $g: \mathbb{R}^+ \to \mathbb{R}_{\geq 0}$, and the functional equation becomes $$g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$$Let $k=2+f(1)$. Picking an arbitrary $t>0$, by $P(t,k)$, we find that $g(t+k)\geq g((c+1)t+k+f(k))$, which really just means that we can always find some $t'\geq(c+1)t$ such that $g(t'+k)\leq g(t+k)$. By repeatedly applying this, we find that there exists some $t''\geq 2t+k$ such that $g(t''+k)\leq g(t+k)$. But then from $P(t''-2t-k,t+k)$, we find that $g((c+1)(t''-2t-k)+g(t+k)+2(t+k))\leq 0$, with equality holding only if $g(t''+k)=g(t+k)=0$. Therefore $g(y)=0$ when $y>k$. But then from $P(k,x)$ where $x>0$ is arbitrary, we find that $g(x)=0$ for all $x$, hence $g \equiv 0$ and we extract the desired solution. $\blacksquare$

The answer is $f(x) = 2x$ only. This works. Notice that equation $(c+1)x+f(y) = x+2y$ can never have solutions in the positive reals, since otherwise $x$ or $c$ would have to be $0$. This implies that $f(y) \ge 2y \phantom{jj} (*)$ for all $y > 0$. We have two cases. Case 1. $f(t) = 2t$ for some $t > 0$. By $*$, we have $$f(x+2y) = f((c+1)x + f(y)) - 2cx \ge 2(c+1)x + 2f(y) -2cx\ge 2x + 4y.$$Pick some $(x,y)$ with $x+2y = t$. Then the above inequality is actually an equality, so we have $f(y) = 2y$ and $f((c+1)x + 2y) = 2(c+1)x + 4y$ for all $x+2y = t$. The second equality simplifies to $f(t + cx) = 2t + 2cx$ for all $x < t$. Since $c+1 > 1$, repeating this process implies that $f(x) = 2x$ for all $x \ge t$, since the function $(c+1)^n$ grows without bound. In particular, $f(2t) = 4t$, so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$. Case 2. $f(t) > 2t$ for all $t>0$. Then let $g(x) = f(x) - 2x$. Clearly the codomain of $g$ is also $\mathbb{R}_{>0}$. The given equation simplifies to $g(x+2y) = 2g(y) + g((c+1)x+g(y) + 2y)$. This implies $g(x+2y) > 2g(y)$, so for all pairs $(x,y)$ with $x>2y$, we have $g(x) > 2g(y)$. Let $g(1) = M>0$. Then for all $x > 2$, $g(x) > 2M$. Taking $y = 1$ into the given equation, \begin{align*} g(x+2) &= 2M + g(\textcolor{blue}{(c+1)x+M}+2) \\ &= 2M + 2M + g((c+1)(\textcolor{blue}{(c+1)x+M}) +M + 2) \\ &= 2M + 2M + 2M + g((c+1)((c+1)(\textcolor{blue}{(c+1)x+M})+M) +M + 2) \\ &= \cdots, \end{align*}which is a contradiction as $M > 0$. We are done.

Kind of easy P4, anyway, as it's FE here's my solution : $~$ Let $P(x,y)$ be the assertion : $$ f((c+1)x+f(y))=f(x+2y)+2cx$$Claim : $f(x)\ge 2x ~\forall x \in \mathbb{R_+}$ Proof : if $\exists w \in \mathbb{R_+}~ ; f(w)<2w :$ $$P(\frac{2w-f(w)}{c},w): f(2w+\frac{2w-f(w)}{c})=f(2w+\frac{2w-f(w)}{c})+2(2w-f(w))$$$\implies f(w)=2w$ contradiction. $\blacksquare$ Define :$$g:\mathbb{R_+} \to \mathbb{R}_{\geq 0} ~,~ g(x)=f(x)-2x\implies f(x)=g(x)+2x$$$\bullet ~P(x,y)$ becomes $:$ $$g((c+1)x+2y+g(y))+2g(y)=g(x+2y)$$$$g((c+1)x+2y+g(y))\ge 0\implies g(x+2y)\ge 2g(y)$$and so $x>2y\implies g(x)\ge 2g(y) $ $\bullet$ Now assume that$: \exists\alpha\in\mathbb{R_+}~ ; g(\alpha)>0$ $$P(x-\frac{2\alpha+g(\alpha)}{c+1},\alpha) ~\text{with}~ x>\frac{2\alpha+g(\alpha)}{c+1} : $$$g((c+1)x)+2g(\alpha)=g(x+\frac{2c\alpha-g(\alpha)}{c+1})$ $$a:=c+1>1,r:=\frac{1}{a}<1,t:=2g(\alpha)>0,d:=\frac{2c\alpha-g(\alpha)}{c+1}\in\mathbb{R}$$$$\implies g(ax)+t=g(x+d)\implies g(ax)=g(x+d)-t$$§ Now change $x\to ax:$ $$g(a^2x)=g(ax+d)-t=g(a(x+rd))-t=g(x+rd+d)-2t$$§ again change $x\to ax:$ $$g(a^3x)=g(ax+rd+d)-2t=g(a(x+r^2d+rd))-2t=g(x+r^2d+rd+d)-3t$$§ By induction we get $:$ $$ {g(a^nx)+nt=g(x+d\frac{1-r^n}{1-r}) ~ \forall n \in\mathbb{N}} $$$\bullet$ Now put $x=x_0$ where $x_0$ is a fixed positive real number satisfying $:$ $x_0+d\frac{1-r^n}{1-r}>0, \forall n\in\mathbb{N}$ which is exist because $\frac{1-r^n}{1-r}<\frac{1}{1-r}~$ (bounded) So $x_0+d\frac{1-r^n}{1-r}$ is bounded $\forall n\in\mathbb{N}$ $$\implies \exists l \in\mathbb{R_+}~ ; l>x_0+d\frac{1-r^n}{1-r} ~\forall n\in\mathbb{N}$$$$\implies 2l>2(x_0+d\frac{1-r^n}{1-r})\implies g(2l)\ge 2g(x_0+d\frac{1-r^n}{1-r})~\forall n\in\mathbb{N}$$So $g(x_0+d\frac{1-r^n}{1-r})$ is bounded $\forall n\in\mathbb{N}$ $\bullet$ Now take $n\to +\infty$ in the equation $:$ $$g(a^nx_0)+nt=g(x_0+d\frac{1-r^n}{1-r})$$we see that $RHS$ is bounded, but $LHS$ goes to infinity, so we get a contradiction and so no such $\alpha$ exist.

See this thread from 2019. The solution should adapt.

The only solution is $f(x)=2x$, which evidently works. From now on, assume that $f$ is a solution which is different from this function. We prove a series of Claims: Claim 1: $f(x) \geq 2x$ for all $x>0$. Proof: Indeed, if $f(u)<2u$ for some $u>0$, then we may take $x=\dfrac{2u-f(u)}{c}$ and $y=u$ in the given equation to obtain that $2c \cdot \dfrac{2u-f(u)}{c}=0,$ a contradiction $\blacksquare$ Claim 2: If $x>2y$, then $f(x)-2x \geq 2(f(y)-2y)$. Proof: Indeed, note that $f(x+2y)-2(x+2y) =f((c+1)x+f(y))-2cx-2x-4y \geq 2((c+1)x+f(y))-2cx-2x-4y=2(f(y)-2y),$ and so $f(x)-2x \geq 2(f(y)-2y)$ for all $x,y$ such that $x>2y$ $\blacksquare$ To the problem, note that $(c+1)x+f(y)>f(y)>2y,$ and so $f((c+1)x+f(y))-2((c+1)x+f(y)) \geq 2(f(y)-2y),$ implying that $(f(x+2y)+2cx)-2((c+1)x+f(y)) \geq 2(f(y)-2y)$. Therefore, $f(x+2y)-2x-2f(y) \geq 2f(y)-4y,$ or equivalently $f(x+2y) \geq 2x+4f(y)-4y$. This readily implies that $f(x+2y)-2(x+2y) \geq 4(f(y)-2y)$, and so $f(x)-2x \geq 4(f(y)-2y)$ for all $x,y>0$ such that $x>2y$. We may generalize this property: Claim 3: If $x,y>0$ such that $x>2^ny$ for some positive integer $n$, then $f(x)-2x \geq 4^n(f(y)-2y)$. Proof: We proceed by induction on $n$. The base case is already established. Now, suppose that $f(x)-2x \geq 4^n(f(y)-2y)$ for all $x,y$ such that $x>2^ny$. Then, if $x>2^{n+1}y$ we may pick an $s$ such that $s \in (2y, \dfrac{x}{2^n})$. Therefore, $x>2^ns$ and $s>2y$. Thus, using the inductive hypothesis and the base case, $f(x)-2x \geq 4^n(f(s)-2s) \geq 4^{n+1}(f(y)-2y),$ as desired $\blacksquare$ To the problem, let $x,y>0$ such that $x>2y$. Let $n=\lfloor \log_2{\dfrac{x}{y}} \rfloor \geq 1$. Then, $x>2^n y$, and so by Claim 3 we infer that $f(x)-2x \geq 4^{n}(f(y)-2y) \geq 4^{\log_2{\dfrac{x}{y}}-1}(f(y)-2y),$ or equivalently that $f(x)-2x \geq \dfrac{x^2(f(y)-2y)}{4y^2}$. Now, take a $y_0$ such that $f(y_0) \neq 2y_0,$ and let $\dfrac{f(y_0)-2y_0}{4y_0^2}=A>0$. Then, $f(x) \geq Ax^2+2x$ for all large $x$. Since $A>0$, we obtain that $f(x)> 4x+1-c$ for all large $x$. Therefore, $(f(y)+c+1)-2(2y+1)=f(y)-4y+c-1 >0,$ if $y$ is large, and so by Claim 2 we obtain that $f(2y+1)+2c=f(c+1+f(y)) \geq 2(c+1+f(y))+2(f(2y+1)-2(2y+1)),$ or equivalently that $f(2y+1)+2f(y) \leq 8y+2,$ which in light of Claim 1 implies that equality must hold, and so $f(x)=2x$ for all large $x$, say $x>N$. Now, for any $y$ we may take a positive integer $n$ such that $2^ny>N$, and so for all $x>2^ny$ we obtain that $f(x)=2x$. By Claim 3, this implies that $0=f(x)-2x \geq 4^n(f(y)-2y) \geq 0,$ and so equality must hold everywhere, thus $f(x)=2x$ for all $x$, which is a contradiction to our initial assumption. To sum up, the only solution is $f(x)=2x$ for all $x>0$.

CyclicISLscelesTrapezoid wrote: Let $c>0$ be a given positive real and $\mathbb{R}_{>0}$ be the set of all positive reals. Find all functions $f \colon \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ such that \[f((c+1)x+f(y))=f(x+2y)+2cx \quad \textrm{for all } x,y \in \mathbb{R}_{>0}.\] $P(x+2z,y)-P(x+2y,z)$ implies that $\frac{(2z(c+1)+f(y))-(2y(c+1)+f(z))}{4c(z-y)}$ is constant (for $y\neq z$). The rest is easy. (The linked lemma can't be applied directly, but the same proof can be used here.)

A standard R+ functional equation lol. Let $P(x, y)$ be the assertion. First, if there exists $y > 0$ with $f(y) < 2y$, there exists $x > 0$ such that $(c+1)x+f(y)=x+2y$. With the original equation, we deduce that $x = 0$ which is a contradiction. Therefore, $\boxed{f(y) \geq 2y, \, \forall y > 0}$. Then we have \[ f(x + 2y) + 2cx = f((c + 1)x + f(y)) \geq 2((c + 1)x + f(y)) \implies f(x + 2y) \geq 2(x + f(y)). \]In particular, $\boxed{f(x + 2y) > f(y)}$ for all $x, y > 0$. Now we will prove the injectivity. Suppose that $f(a) = f(b)$ for some $a > b > 0$. Compare with $P(x, a)$ and $P(x, b)$ we know that $f(2a + x) =f(2b + x)$. It is a periodic function with $f(2b) = f(2b + (2a - 2b) \cdot k)$ for all $k \in \mathbb{N}$. We can choose $k$ sufficiently large so that $2b + (2a - 2b) \cdot k > 2 \cdot 2b$. Combined with the previous inequality, it is a contradiction. Hence $f$ is injective. Next, compared with $P(\frac{z}{c+1} + 2x, y)$ and $P(\frac{z}{c + 1} + 2y, x)$ we have \[ f(2(c + 1)x + f(y) + z) - f(2(c + 1)y + f(x) + z) = 4(x - y). \]Also, compared with $P(2x, y)$ and $P(2y, x)$, the above equation also holds for $z = 0$. For given $x, y > 0$, we select $t > x, y$, then \[ f(2(c + 1)(x + t) + f(y + t)) - f(2(c + 1)(y + t) + f(x + t)) = 4(x - y). \]Since $f(x + t) > f(x)$ and $2(c + 1)(y + t) > 2(c + 1)y$, there exists $z > 0$ with \[2(c + 1)y + f(x) + z = 2(c + 1)(y + t) + f(x + t).\]Compared with the previous two equations and by the injectivity of $f$ we get \[2(c + 1)x + f(y) + z = 2(c + 1)(x + t) + f(y + t).\]Thus we have \[ f(x) - f(y) = f(x + t) - f(y + t). \]The equation also holds for all $t > 0$ since for given $x, y, z > 0$ we can choose a $A > \max\{x + 2z, y + 2z\}$, then \[ f(x) - f(y) = f(x + A) - f(y + A) = f((x + z) + (A - z)) - f((y + z) + (A - z)) = f(x + z) - f(y + z) \] With this strong equation, we can substitute the original equation into \[ Q(x, y): f(cx + f(y)) = f(2y) + 2cx \implies f(x + f(y)) = f(2y) + 2x \]Compared with $Q(f(x), y)$ and $Q(f(y), x)$ we have $f(2y) = 2f(y)$, so \[ Q(x, y) \implies f(x + f(y)) = 2(x + f(y)), \]which also implies there exists a constant $M$ such that $f(t) = 2t$ for all $t > M$. Finally, for all $x > 0$, there exists $k \in \mathbb{N}$ such that $2^k x > M$, then we have $f(x) = \frac{f(2^kx)}{2^k} = \frac{2 \cdot 2^kx}{2^k} = 2x$. Hence the unique solution is $\boxed{f(x) \equiv 2x}$.

Let $P(x,y)$ be the given assertion $(c+1)x+f(y)\neq x+2y\Rightarrow 2y-f(y)\leq 0\Leftrightarrow f(x)\geq 2x,x>0$ Define $g:\mathbb{R}_{>0}\to \mathbb{R}_{\geq 0}$, $ g(x)=f(x)-2x$ Now $P(x,y)$ becomes $Q(x,y): 2g(y)+g((x+1)c+f(y))=g(x+2y)$. $Q(x,y)\Rightarrow g(x)\geq 2g(y)$ whenever $x> 2y$ Now $(x+1)c+f(y)>2y$ so $g((x+1)c+f(y))\geq 2g(y)$ and thus $Q(x,y)\Rightarrow g(x)\geq 4g(y),x> 2y$. Continuing in the same way we find that $g(x)>kg(y)$, when $x>2y$ for arbitrary large $k$'s and thus $g(y)=0$ etc

Let $P(x,y)$ denote the assertion. If $f(y)<2y$ for some $y$ then $P\left(x,\frac{2y-f(y)}{c}\right)$ lead the contracdition. So $f(x)\ge 2x\quad\forall x\in\mathbb{R}^+ \ (1)$. $$P(x,y)\implies f(x+2y)+2cx\ge 2(c+1)x+2f(y)\iff f(x+2y)\ge 2f(y)+2x\quad\forall x,y\in\mathbb{R}^+.$$Let denote this assertion is $Q(x,y)$. There are two cases, Let $V= \{x\in\mathbb{R^+}| \ f(x)=2x\}$. $\textbf{Case 1:}$ $V \ \text{is empty}$. We obviously have $f(x)>2x$ for $x>0$. Pick any $u$ such that $g(u)=f(u)-2u=A>0$. $$P(x-2u,u)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu,\quad\forall x>2u.$$$$Q\left(2cx-4u,x\right)\implies f(cx+x-2uc+A)=f(x)+2cx-4cu\le \frac{f\left(2x+2cx-4cu\right)}{2},\quad\forall x>2u.$$Let $t=cx+x-2uc$ implies that $f(t+A)\le \frac{f(2t)}{2} \quad\forall t>2u. \ (2)$ If $A=f(u)-2u>2u$ then let $t=A$ leads the contracdition, so $f(u)\le 4u$. We claim that if $f(x)>2x$ then $f(x)\le 4x$. And so in this case $2x<f(x)\le 4x \ \forall x\in\mathbb{R}^+.$ And so $\lim_{x\to 0^+} f(x)=0$. We use $(2)$ to have $f(2t)\ge 2f(t+A)> 4t+4A \implies f(t)\ge 2t+4A\quad\forall t>4u.$ Let $y>2u$, since $f(y)>2y>4u$, then by $P(x,y)\implies f(x+2y)+2cx> 2(c+1)x+2f(y)+4A\implies f(x+2y)>2x+2f(y)+4A \ \quad\forall x>0, y>2u. \ (99)$ Obvious, we also have $$4x+8y\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u.$$Let $x\to 0$, then $f(y)\le 4y-2A\quad\forall y>2u.$ By $Q(x,y)$, we have $$4x+8y-2A\ge f(x+2y)>2x+2f(y)+4A\quad\forall x>0, y>2u. \ (100)$$Let $x\to 0$ again, we get that $f(y)\le 4y-3A\quad\forall y>2u$, we can do similarly $(100)$ to obtain that $f(y)\le 4y-4A, \ \forall y>2u$. And so $2x+2f(y) \le f(x+2y)\le 4x+8y-4(f(y)-2y)=4x+16y-4f(y)\quad\forall x>0, y>0. \ (101)$ Let $x\to 0$, we get that $f(y)\le \frac{8}{3}y, \ y>0$. By $(99)$ we have $$2x+2f(y)+4A<f(x+2y)\le \frac{8}{3}(x+2y)\quad\forall x>0, y>2u.\ (102)$$Let $x\to 0$ to see that $f(y)\le \frac{8}{3}y-2A\quad\forall y>2u $, do similarly $(100)$ we have $f(y)\le\frac{8}{3}y-4A$ forall $y>2u$. From this clause, similarly $(102)$, which have $$2x+2f(y)\le f(x+2y)\le \frac{8}{3}(x+2y)-4(f(y)-2y)\quad\forall x,y>0.$$We can do similarly each step in $(100),(101),(102)$ to see that if $f(x)\le v_nx \ \forall x>0$ then $f(x)\le v_{n+1}x$ which sequence $(v_n)$ is determined by $$\begin{cases} \displaystyle \ v_1=4 \\ \ \displaystyle v_{n+1}=\frac{v_n+4}{3}. \end{cases}$$It's easy to see that $(v_n)$ is decreasing and $\lim v_n=2$, and so $2x< f \le 2x\quad\forall x>0$, which is asburd. $\textbf{Case 2:} \ \text{there exists} \ a\in V$ $$Q\left(x,\frac{a-x}{2}\right)\implies 2a=f(a)\ge 2f\left(\frac{a-x}{2}\right)+2x\quad\forall 0<x<a.\implies f\left(\frac{a-x}{2}\right)\le a-x\implies f\left(\frac{a-x}{2}\right)=a-x.\ \forall 0<x<a.$$Which means that $f(x)=2x\quad\forall 0<x<\frac{a}{2}.$ $$P\left(x,\frac{a}{8}\right)\implies f\left((c+1)x+\frac{a}{4}\right)=2cx+f\left(x+\frac{a}{4}\right),\quad\forall x>0.$$Let $x\in\left(0,\frac{a}{4}\right)$ implies that $f\left((c+1)x+\frac{a}{4}\right)=2cx+2x+\frac{a}{2}\quad\forall \ 0<x<\frac{a}{4}$. Let $y=(c+1)x+\frac{a}{4}$ then $f(y)=2y\quad\forall y\in\left(\frac{a}{4},\frac{ca}{4}+\frac{a}{2}\right) \ (3)$. Denote the sequence $(x_n)$ $$\begin{cases} \displaystyle \ x_1=\frac{a}{2} \\ \ \displaystyle x_{n+1}=\left(x_n-\frac{a}{4}\right)(c+1)+\frac{a}{4} \ \end{cases} \quad\forall n\in\mathbb{Z}^+.$$It's easy to see that $x_n=\frac{a}{4}(c+1)^{n-1}+\frac{a}{4}$, hence $\lim x_n=+\infty$, and use induction to see that $f(x)=2x\quad\forall x\in \left(\frac{a}{4},x_n\right)$ (continue the process similarly $(3)$ is not hard), but since $f=2x$ in $\left(0,\frac{a}{2}\right)$ so $f(x)=2x\quad\forall x\in (0,x_n)$. Thus, the only sol is $$\boxed{f(x)=2x\quad\forall x\in\mathbb{R}^+}.$$

Let $P(x, y)$ be the assertion above. $\textbf{Claim: } f(x) \geq 2x$ $\textit{Proof}$ If we assume contrary, we can have $P\left(\frac{2y-f(y)}{c}, y\right)$ to reach a contradiction. $\square$ Now define $g: \mathbb{R^+} \rightarrow \mathbb{R}_{\geq 0}, g(x) := f(x) - 2x$. We get that \[g((c+1)x + g(y)+2y)+2g(y) = g(x+2y)\]and redefine $P(x,y)$ for this equation. Notice immediately we get $2g(y) \le g(x+2y)$. As $g$ is bounded on intervals, by Bolzano-Weirstrass theorem, considering a converging sequence $x_n \rightarrow 0$ such that $y = x_i, x+2y = x_j$ gives us that $g(x_n) \rightarrow 0$. As $2g(y) \le g(x+2y)$, we can easily see that every sequence limiting to $0$ gives $g \rightarrow 0$, i.e $\lim_{x \rightarrow 0} g(x) = 0$. Now taking only $y \rightarrow 0$ gives us that $g(Nx) \leq g(x)$ where $N$ can be values arbitrarily close to $c+1$. This implies $g(Mx) \leq g(x)$ for values $M$ arbitrarily close to $(c+1)^n, \forall n \in \mathbb{N}$. Now then having $\frac{g(z+2x)}{2} \geq g(x) \geq g(Mx)$, letting $z = Mx-2x$ for a sufficiently large $n$ gives $g \equiv 0 \rightarrow f \equiv 2x$.

Let $P(x,y)$ be the original equation. If $f(y)<2y$ for some $y\in\mathbb{R}^+$, then $$P\left(\dfrac{2y-f(y)}{c},y\right)\implies 2(2y-f(y))=0\ (\text{contradict}).$$So we have $f(y)\geq 2y$. If $f(a)=f(b)$ for some $a-b>0$. Comparing $P(x,a)$ and $P(x,b)$, $$f(x+d)=f(x)\quad\forall x\gg 0,$$where $d=2(a-b)$. Take $n$ sufficiently large such that exists $x_0>0$ s.t. $$(c+1)x_0+f(1)=x_0+2,$$then $P(x_0,1)\implies 2cx_0=0$ (contradict). Hence $f$ is injective. Consider \[f((c+1)x+f(y))+2cz=f(x+2y)+2c(x+z),\]and apply the original equation on both sides, we have $$f\left((c+1)z+f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)\right)=f\left((c+1)(x+z)+f\left(y-\dfrac{z}{2}\right)\right),$$for $2y>z$ (this implies $f(y)\geq 2y>z$). By injective, we have $$f\left(\dfrac{(c+1)x+f(y)-z}{2}\right)=(c+1)x+f\left(y-\dfrac{z}{2}\right).$$Compare above equation with $P\left(\dfrac{(c+1)x}{2c},\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right)$ and by injective, we have \[\dfrac{(c+1)x+f(y)-z}{2}=\dfrac{(c+1)^2x}{2c}+f\left(\dfrac{1}{2}\left(y-\dfrac{z}{2}-\dfrac{(c+1)x}{2}\right)\right).\]Let $y=2w+\dfrac{z+(c+1)x}{2}$, we have $$f\left(2w+\dfrac{z+(c+1)x}{2}\right)=2f(w)+(c+1)x+z \quad \forall x,z>0,$$that is $f(x+2w)=2f(w)+2x$ $\forall w,x\in\mathbb{R}^+$, denoted this equation by $Q(x,w)$. Compare $Q(2x,w)$ and $Q(2w,x)$, we get $f(x)=2x$ for all $x\in\mathbb{R}^+$.

Let $P(x,y)$ be the given functional equation. Obviously $f(x)=2x$ is a solution. Let's prove that it is the unique one. Claim 1. $f(x)\geq 2x$. Proof. Assume there is $x_{0}$ such that $f(x_{0})<2x_{0}$. $P(\frac{2x_{0}-f(x_{0})}{c}, x_{0})$: $2(2x_{0}-f(x_{0}))=0$ - absurd. The claim follows. Now, let $g(x)=f(x)-2x$ and rewrite $P(x,y)$ in terms of $g$. We get the following $$g((c+1)x+2y+g(y))+2g(y)=g(x+2y).$$We'll prove that $g \equiv 0$ which will prove the desired result. Assume there is $y_{0} \in \mathbb{R^{+}}$ such that $g(y_{0})>0$ (the fact that $f(x)\geq 2x$ means that $g(x) \geq 0$). In the last assertion we take $x=p-2y_{0}, y=y_{0}$ to get that: $$g((c+1)p-2y_{0}c+g(y_{0}))+2g(y_{0})=g(p), \forall p>2y_{0}; \longrightarrow Q(p)$$Now, lets fix $p>2y_{0}$ and define the following sequence $\textbf{a} =\{a_{i}\}_{i=0}^{\infty}: a_{0}=p, a_{n}=(c+1)a_{n-1}+g(y_{0})-2cy_{0}$. We see that $a_{n}-a_{n-1}=c(a_{n-1}-2y_{0})+g(y_{0})$. Because of $a_{0}>2y_{0}$ by induction we get that $\textbf{a}$ is strictly increasing, so $a_{n}>2{y_0}$. It means that $g(a_{i+1})+2g(y_{0})=g(a_{i})$ from $Q(a_{i})$. Summing the last one for $i=0,1,...,n-1$ we get that $g(a_{n})+2ng(y_{0})=g(a_{0})$, so $g(a_{0})\geq2ng(y_{0})$ which contradicts the fact $n$ can tend to infinity. So, $g\equiv 0$ and we are done.

Let $P(x,y)$ be the denotation of the given equation. Lemma: The function $f$ is injective. Proof: Assume not. Then there exists $a<b$ such that $f(a)=f(b)$. From comparing $P(x,a)$ and $P(x,b)$ we obtain $$f(x+2a)=f(x+2b)$$In other words $f(x)=f(x+2(b-a))$ for all $x>2a$. Thus $f$ is periodic on $(2a,\infty)$. Let the smallest period be $T$. From comparing $P(x,y)$ and $P(x+\frac{T}{c+1},y)$ we obtain $f(x+2y)=f(x+\frac{T}{c+1}+2y)+\frac{2cT}{c+1}$ and with induction if $z=x+2y$ then $$f(z)=f(z+\frac{nT}{c+1})+\frac{2cnT}{c+1}$$for all positive integer $n$. But if $n>\frac{(c+1)f(z)}{2cT}$ we'll get a contradiction since the right-hand side is obviously bigger than the left-hand side. Thus the conclusion follows. From compairing $P((c+1)x,(c+1)^2x+f(y))$ and $P((3c+1)x,(c+1)x+2y)$ we get the equation $$f((5c+3)x+4y)+4c^2x=f((2c^2+5c+3)x+2f(y))..........(1)$$From equation $(1)$, $P(2cx,2y+\frac{3(c+1)}{2}x)$ and injectivity we get $f(2y+\frac{3(c+1)}{2}x)=3(c+1)x+2f(y)$. And with rewriting $\frac{3(c+1)}{2}x$ as $x$ we obtain $$f(x+2y)=2x+2f(y)$$After writing this equation on the original equation and rewriting $(c+1)x$ as $x$ we get $$f(x+f(y))=2x+2f(y)=f(x+2y)$$With injectivity, we obtain $f(x)=2x$ for all positive reals $x$ and obviously, this function works on the original equation. Thus the only function that satisfies this equation is $f(x)=2x$ for all positive reals $x$

We uploaded our solution https://calimath.org/pdf/APMO2023-4.pdf on youtube https://youtu.be/yRc3q7caBvk.

Stupid Problem, The only solution is $f(x)=2x$ Claim 1:$f(x) \geq 2x$

Now the idea is to define $g(x)=f(x)-2x$, the equation now becomes $g((c+1)x+g(y)+2y)+2g(y)=g(x+2y).$ Claim 2 $g \equiv 0$

Hence we're done

kred9 wrote: In particular, $f(2t) = 4t$, so we have $f(y) = 2y$ for all $y < t$ as well, showing that $f(x) = 2x$ for all $x \in \mathbb{R}_{>0}$. I don’t see why $f(2t) = 4t$ implies that $f(y) = 2y$ for all $y < t$. Could you explain this to me?

Looks like all the solutions have very alike ideas, however everybody explains them differently, I will take a try. First, denote the assertion by $P(x;y)$, assuming there exists $y$ such that $f(y)<2y$, from $P(\frac{2y-f(y)}{c};y)$ we get a contradiction. Assume there exists some $y_1$ such that $f(y_1) \neq 2y_1$, I will achieve a contradiction. First, apply $f(a) \ge a$ to the $LHS$ to obtain after reducing that $2x+2f(y) \le f(x+2y)$. Now define $g:\mathbb{R_+} \rightarrow \mathbb{R}_{\ge0}$, $g(x)=f(x)-2x$. The given property reduces to $g(x+2y) \ge 2g(y)$, call it $Q(x;y)$. First, I will prove $g$ is unbounded. It is not hard, as $Q(x;x) \Rightarrow g(3x) \ge 2g(x)$, and iterating for $y_1$, we achieve the desired result. Now I will prove that for any given constant $k$ there exists another constant $N_k$ such that $g(x) \ge kx, \forall x \ge N_k$. For this just consider $Q(\epsilon y;y)$ and iterate to get $g((2+\epsilon)^ny_0) \ge 2^ng(y_0)$. Take a sufficiently large $g(y_0)$. Considering the largest $n$ such that $2^ny_0 <x$ , and taking $\epsilon$ such that $(2+\epsilon)^ny_0=x$, I get of course together with $2^ny_0 \ge \frac{x}{2}$, that for $x \ge 2y_0$, $g(x) \ge \frac{x}{2}g(y_0)$, which concludes the claim. From here, just pick a large enough $k$ and $x \ge N_k$, and applying the inequality $f(x) \ge (2+k)x$ to the $LHS$ of $P(x;y)$, together with small iterations shows that for all big enough $x$, $f(x)$ is unbounded, which obviously can't be true. This is a very good example of FE, where you use the most standart $\mathbb{R}_+ \rightarrow \mathbb{R}_+$ techniques.

I hope there is nothing wrong . The only function is $f(x)=2x \ \forall \ x \in \mathbb{R^+}$. Easy to check that this satisfies, now we'll prove that this is the only one. Let $P(x,y)$ be the assertion. All variables mentioned should be in $\mathbb{R^+}$ unless specified. Claim 1. $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$ Proof. Suppose there exist $k \in \mathbb{R^+}$ such that $f(k)<2k$. Then $P(\frac{2k-f(k)}{c},k)$ gives $$f\left( \frac{2k-f(k)}{c}+f(k) \right)=f\left( \frac{2k-f(k)}{c}+f(k) \right) +2c\left( \frac{2k-f(k)}{c} \right)$$Clearly this isn't possible since $2c\left( \frac{2k-f(k)}{c} \right) >0$. As desired. Claim 2. $f$ is injective Proof. Suppose there exist distinct $a,b \in \mathbb{R^+}$ such that $f(a)=f(b)$, we may assume $a>b$. Comparing $P(x,a)$ and $P(x,b)$ yields $$f(x+2a)=f(x+2b) \Leftrightarrow f(x)=f(x+(2a-2b)) \ \forall \ x > 2a$$. By Induction, we have $$f(x)=f(x+n(2a-2b)) \ \forall \ x > 2a $$for all $n \in \mathbb{N}$. Note that $x+n(2a-2b)$ goes to infinity as $n$ goes to infinity. Since $f(x)\ge 2x \ \forall \ x \in \mathbb{R^+}$, $f(x)$ is infinity for all $x>2a$. Clearly this is absurd because technically infinity is not a real number. So this is a contradiction. As desired. Claim 3. $f(x)-2x$ is constant Proof. From $P\left( x+y,\frac{z}{2} \right)$, we have $$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f(x+y+z)+2cx+2cy \ (1)$$. From $P\left( x,\frac{y+z}{2} \right)$, we have $$f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)=f(x+y+z)+2cx \ (2)$$. Plugging $(2)$ to $(1)$ yields $$f\left( (c+1)(x+y)+f\left( \frac{z}{2} \right) \right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right)+2cy \ $$. Note that $\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}>0$. From $P\left( y, \frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2}\right)$, we have $$f\left( (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)\right)=f\left( (c+1)x+f\left( \frac{y+z}{2} \right) \right) + 2cy$$. Injectivity gives $$(c+1)(x+y)+f\left( \frac{z}{2} \right)= (c+1)y + f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$$$$\Leftrightarrow (c+1)x+f\left( \frac{z}{2} \right)=f\left(\frac{(c+1)x+f\left( \frac{y+z}{2} \right)-y}{2} \right)$$. Change $x$ to $\frac{y}{c+1}$ to get $$y+f\left( \frac{z}{2} \right)=f\left(\frac{f\left( \frac{y+z}{2} \right)}{2} \right)$$. The RHS is symmetric in $y$ and $z$, so swapping them yields $$y+f\left( \frac{z}{2} \right)=z+f\left( \frac{y}{2} \right) \Leftrightarrow f\left( \frac{y}{2} \right)-y \ is \ constant$$. Just change $y$ to $2y$ and we are done. Now, plug $f(x)=2x + c \ \forall \ x \in \mathbb{R^+}$ for a constant $c \in \mathbb{R}$ to $P(x,y)$. Easy to check that $c=0$ by pairing the coefficients. As desired.