= Solution
For $x=\cos\theta$ with $0<\theta<\pi$, differentiation of the <Chebyshev polynomial> gives
$$
T_n'(\cos\theta)=n\frac{\sin(n\theta)}{\sin\theta},\qquad \frac{1-x^2}{n^2}T_n'(x)^2=\sin^2(n\theta)=1-T_n(x)^2.
$$
Both sides are <polynomials> in $x$, so the identity holds at the endpoints and, in fact, for every real $x$.
Order the <roots of a polynomial> here as $x_j=\cos\theta_j$, $\theta_j=(2j-1)\pi/(2n)$, so $x_1>\cdots>x_n$. At these simple roots,
$$
T_n'(x_j)=\frac{n(-1)^{j-1}}{\sin\theta_j},\qquad |T_n'(x_j)|=\frac{n}{\sqrt{1-x_j^2}}.
$$
For a <polynomial> $q$ of degree at most $n-1$, the <Lagrange interpolation polynomial> is
$$
q(x)=\sum_{j=1}^n q(x_j)\ell_j(x),\qquad \ell_j(x)=\frac{T_n(x)}{(x-x_j)T_n'(x_j)}.
$$
For $x>x_1$, the <Chebyshev polynomial> is positive: all its roots are at most $x_1$ and its leading coefficient is positive. Hence $\ell_j(x)T_n'(x_j)=T_n(x)/(x-x_j)>0$. The assumed nodal bounds and interpolation of the degree-$n-1$ <polynomial> $T_n'$ now give
$$
|q(x)|\le\sum_j|\ell_j(x)|\,|T_n'(x_j)|=\sum_j\ell_j(x)T_n'(x_j)=T_n'(x).
$$
Continuity includes $x=x_1$. Apply this argument to $q(-x)$ and use the parity of the <Chebyshev polynomial> to obtain the negative side. Thus the <Chebyshev nodal derivative comparison> proves
$$
\boxed{|q(x)|\le|T_n'(x)|\quad\text{when }|x|\ge\cos(\pi/(2n)).}
$$
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