Chebyshev polynomials increase more quickly than any other polynomial outside $[-1,1]$ In Appendix C3 of Shewchuk's excellent notes on conjugate gradient, it is stated without proof that

Chebyshev polynomials... increase in magnitude more quickly outside the range $[-1,1]$ than any other polynomial that is restricted to have magnitude no greater than one inside the range $[-1,1]$.

Here, we mean Chebyshev polynomials of the first kind, $T_n(x)=\cos(n \cos^{-1} x)$ for $x \in [-1,1]$.   I have not been able to find a proof of this fact anywhere, and don't even know what I would search for in the first place.  I have tried proving it myself, supposing that some polynomial attains a larger magnitude and attempting to arrive at a contradiction by showing, for example, that such a polynomial has too many zeros.  
How might I begin to prove something like this / where can I find more information?
 A: Recently, I need this result for another question. Since I cannot find a easy reference of this, I have proved it myself. I will extract the part relevant to this question and presented here.

Let $f(x)$ be any polynomial with degree at most $n$ and $|f(x)| \le 1$ over $[-1,1]$. 
Let $x_k = \cos\frac{k\pi}{n}$ for $0 \le k \le n$, we have
$1 = x_0 > x_1 > \ldots > x_n = -1$.
For any $\epsilon > 0$, consider the polynomial
$$g(x) = (1+\epsilon)T_{n}(x) - f(x)$$
Since $T_n(x_k) = (-1)^k$, we find: 
$\displaystyle\quad g(x_k) \begin{cases} > 0, & k \text{ even }\\
< 0, & k \text{ odd }\end{cases}$
This implies $g(x)$ has at least $n$ roots $y_1,\ldots,y_n$ with the $k^{th}$ root $y_k$ falls inside the $k^{th}$ interval $(x_k,x_{k-1})$. Since $g(x)$ is a non-zero polynomial with degree at most $n$. These are all the roots of $g(x)$. We can factorize $g(x)$ as $A\prod_{k=1}^n (x - y_k)$ for some constant $A$.
Notice $g(x_0) > 0$ and $x_0 > y_1,\ldots, y_k$. we get $A > 0$. So for all $t \ge x_0 = 1$, we have
$$(1+\epsilon)T_n(t) - f(t) = A\prod_{k=1}^n(t - y_k) > 0 \implies
(1 + \epsilon)T_n(t) > f(t)$$
Apply a similar argument to $-f(x)$, we get
$(1 + \epsilon)T_n(t) > -f(t)$
Combine these, we find $(1 + \epsilon)T_n(t) > |f(t)|$. 
Since this is true for all $\epsilon > 0$, we can deduce $T_n(t) \ge |f(t)|$.
The case for $t \le -1$ is similar. At the end, we have following bound over $\mathbb{R}$.
$$|f(x)| \le \max( 1, |T_n(x)| )$$
