Be $f:\mathbb R^ +\mapsto\mathbb R$ a function that satisfies the following conditions:
a)$ f(f(f(x)))+2x=f(3x)$ for every $x\gt 0$;
b) $\lim_{x \to \infty} (f(x)-x)=0$.
This was proposed by Gabriel Dospinescu at a romanian competitions about a decade ago. No idea how to approach it. Thank you!
The function must be defined as follow $f:\mathbb{R^+}\to\mathbb{R^+}$,otherwise the functional equation can not be true because it involves $f(f(x))$
we know that if replacing $x$ by $\frac{x}{3}$ we obtain : $$f(x)=\frac{2}{3}x+f(f(f(\frac{x}{3})))\geq\frac{2}{3}x$$
now let $u_0=\frac{2}{3}$ assuming that $f(x)\geq u_nx$ one can prove that: $$f(x)\geq u_{n+1}x$$ with $u_{n+1}=\frac{u_n^3+2}{3}$ and because $u_n \rightarrow 1$ when $n$ tends to infinity we conclude that $$f(x)\geq x$$ for evry positive real $x$. Now for any real $a$ let $f(a)=a+t$ then $(3a)\geq 3a+t$ and by induction $f(3^na)\geq 3^na+t$ and using the second condition we have $t=0$
Conclusion The only function verifying the two conditions is the identity.