Cute problem. Let $H(x) = F(x) - G(x)$. We will prove that $H(x) > 0$ for all real $x$. Claim 01. $H$ has no real root. Proof. Suppose otherwise, that $F(a) = G(a)$ for some $a \in \mathbb{R}$, then $F(F(a)) = F(G(a))$, which is a contradiction. By Intermediate Value Theorem, we conclude that either $H(x) < 0$ for all real $x$ or $H(x) > 0$ for all real $x$. Suppose $H(x) < 0$ for all real $x$. This means that $H(G(x)) < 0 \implies F(G(x)) < G(G(x))$, which is a contradiction.

Solution:

Doru2718 wrote: Solution: The statement is true for any continuos functions $F$ and $G$. Take $H=F-G$.If there exists $x_0\in\mathbb{R}$ for which $H(x_0)=0$ then $F(F(x_0))=F(G(x_0))$,contradiction. As $H$ is continuos it has the intermediate value property,so either $H(x)>0$ or $H(x)<0$ for all real $x$. But, from the hyphotesis, $H(G(0))>0$ so $H(x)>0\Rightarrow F(x)>G(x)$,for all real $x$ $\square$ Nice solution

Doru2718 wrote: Polynomials $f$ and $g$ satisfy: $$f(f(x))>g(f(x))>g(g(x))$$for all real $x$.Prove that $f(x)>g(x)$ for all real $x$. Let $h=f-g$. Note that if $f(k)=g(k)$ for some $k$, then $g(g(k))=g(f(k))>g(g(k))$, contradiction. Then by IVT, assume FTSOC that $f(x)<g(x)$ for all $x$. Plugging $x\mapsto f(x)$, we have $f(f(x))<g(f(x))$, contradiction, so $h$ must be all-positive, hence the result.