Iranian Mathematical Olympiad

Let $h(x)=f(x)-x$, so we have $$h(h(x)+g(x)+x+2y)=2h(y).(1)$$ Hence for every $x,y,z\in {\mathbb{R}}^{+}$ we have $$h(h(x)+g(x)+x+2y)=2h(y)=h(h(z)+g(z)+z+2y).(2)$$ There exist some $x,z\in {\mathbb{R}}^{+}$ such that $T=h(x)+g(x)+x-h(z)-g(z)-z$ is positive. Otherwise $h(x)+g(x)+x$ must be constant. So $$h(x)+g(x)+x=h(1)+g(1)+1\Rightarrow h(x)=-g(x)-x+C,$$ where $C=h(1)+g(1)+1$ (constant). By definition of $h(x)$ we get $$f(x)=-g(x)+C.$$ But coefficients of $g(x)$ all are positive, therefore $g(x)\to +\infty$ as $x\to +\infty$ so $g(x)>C$ for large values $x$, which contradicts the assumption that $f(x)\in {\mathbb{R}}^{+}$ for every $x\in {\mathbb{R}}^{+}$. Therefore function $h(x)$ is periodic because according to equation (2) for each $x,z\in {\mathbb{R}}^{+}$, $T=h(x)+g(x)+x-h(z)-g(z)-z$ is a period for $h$. (We can choose $x,z\in {\mathbb{R}}^{+}$ such that $T>0$.) Therefore $$S(x)=h(x+T)+g(x+T)+(x+T)-h(x)-g(x)-x=g(x+T)-g(x)-T.$$ $S(x)$ is a period for $h$ for every $x\in {\mathbb{R}}^{+}$. Since $g(x)$ is a polynomial of degree at least two, $S(x)$ is not constant and is a function of $x$ such that its image contains all numbers greater than a fixed positive real number $A$ and this implies that $h(x)$ is constant for every $x\in {\mathbb{R}}^{+}$ and so $h(x)=K$ for some constant $K$. Now we replace function $h$ by $K$ in (1) $$\begin{gathered} h(h(x)+g(x)+x+2y)=2h(y)\Rightarrow K=2K\Rightarrow K=0 \\ \Rightarrow h(x)=0\forall x\in {\mathbb{R}}^{+}\Rightarrow f(x)=x\forall x\in {\mathbb{R}}^{+}. \end{gathered}$$