Let $f(x)=ax^2 + bx + c$. Suppose $f(x)=x$ has no real roots. Show that the equation $f(f(x))=x$ has also no real solutions.
One of the solution found on this site goes like this:
Since $f$ is continuous, saying that $f(x) = x$ has no solution means that either $f(x) < x$ for all $x$ or $f(x) > x$ for all $x$. Let's assume wlog that $f(x) > x$, then $f(f(x)) > f(x) > x$ for all $x$ so $f(f(x))=x$ has no solution either.
My doubt is, how did they conclude $f(f(x)) > f(x) > x$ from $f(x) > x$ ? Because $a> b \implies f(a)>f(b)$ only when $f$ is strictly increasing, but here we have a quadratic function which is of course not strictly increasing.