Consider function $f,g: I\to I$ where $I$ si compact interval. The function $f$ can be chosen with probability $p$ and the function $g$ can be chosen with probability $1-p$, where $p\in\left(0,1\right)$. Moreover consider that $$ \sup\limits_{x,y\in I;\text{ }x\neq y}\left(\frac{|f(x)-f(y)|}{|x-y|}p+\frac{|g(x)-g(y)|}{|x-y|}(1-p)\right)=c, $$ where $c<1$.
I would like to prove, that $$ \small \sup\limits_{x,y\in I;\text{ }x\neq y}\left(\frac{|f(f(x))-f(f(y))|}{|x-y|}p^2+\frac{|f(g(x))-f(g(y))|}{|x-y|}p(1-p)+\frac{|g(f(x))-g(f(y))|}{|x-y|}(1-p)p+\frac{|g(g(x))-g(g(y))|}{|x-y|}(1-p)^2\right)<c^2. $$
First of all I have been trying to find some counterexample. Then I have been trying to prove it. I am not successful. Honestly, I still thing that it is not true, but I cannot handle with it.
Any help will be appreciated. Thank you very much.