Let $f$, $g$ and $h$ be functions such that $f(x) \leqslant g(x) \leqslant h(x)$ for all $x$ in an open interval containing $a$. Suppose that the curves $y = f(x)$ and $y = h(x)$ have a common tangent line $L$ at $x = a$. Prove that $L$ is also the tangent line to $y = g(x)$ at $x = a$.
Firstly, I let$$\frac{f(x)-f(a)}{x-a}\leqslant\frac{g(x)-g(a)}{x-a}\leqslant\frac{h(x)-h(a)}{x-a}.$$
But how do I prove $\displaystyle\left|\frac{f(x)-f(a)}{x-a}\right|\leqslant\left|\frac{g(x)-g(a)}{x-a}\right| \leqslant\left|\frac{h(x)-h(a)}{x-a}\right|$?
What you want to prove is that, if $f'(a)=h'(a)$, then$$\tag{1}f'(a)=g'(a)=h'(a).$$You know that, if $x>a$, then$$\frac{f(x)-f(a)}{x-a}\leqslant\frac{g(x)-g(a)}{x-a}\leqslant\frac{h(x)-h(a)}{x-a}.$$Therefore by the squeeze theorem, the right derivative of $g$ at $a$ is equal to $f'(a)=h'(a)$. If $x<a$, then what you know is that$$\frac{f(x)-f(a)}{x-a}\geqslant\frac{g(x)-g(a)}{x-a}\geqslant\frac{h(x)-h(a)}{x-a}$$and so, again by the squeeze theorem, the left derivative of $g$ at $a$ is equal to $f'(a)=h'(a)$. Therefore, $(1)$ holds.