Let $T_n$ be the space of all real-valued trigonometric polynomials on $[0,1)$ of degree at most $n\in \mathbb{N}_0$ and $p\in T_n.$ Then $$ \left|p'(x)\right|\leq 2\pi n\sqrt{\|p\|^2_\infty -|p(x)|^2}. \label{1}\tag{$*$} $$ The author proves this by contradiction. He assumes that there is a point $x_0$ and a $p\in T_n$ such that $\|p\|_\infty <1$ and $$ \left|p'(x_0)\right|= 2\pi n\sqrt{1 -|p(x_0)|^2}.\label{2}\tag{$**$}$$ The proof uses mainly the condition of the norm.
My question isn't about the proof, rather about the logic. How does the fact that there are no such functions \eqref{2} imply the required inequality \eqref{1}?
What can be deduced from the contradiction is that for all $p\in T_n$, $$\lVert p\rVert_\infty \geq 1 \text{ or } \forall x\in[0,1), |p'(x)|\neq2\pi n\sqrt{1-|p(x)|^2}$$
Let $p\in T_n$ ($p\neq 0$) and $k\in\left[0,\frac{1}{\lVert p\rVert_\infty}\right)$. We get $\lVert kp\rVert_\infty<1$. Therefore, for $x\in[0,1)$, $$f(k,x):=|kp'(x)|-2\pi n\sqrt{1-|kp(x)|^2}\neq 0$$ $f(\cdot,x)$ is continuous, and $f(0,x)=-2\pi n<0$ thus by intermediate value theorem, $f(k,x)<0$ for $k\in\left[0,\frac{1}{\lVert p\rVert_\infty}\right)$. Having $k\to\frac{1}{\lVert p\rVert_\infty}$ yields $$\left|\frac{p'(x)}{\lVert p\rVert_\infty}\right|-2\pi n\sqrt{1-\left|\frac{p(x)}{\lVert p\rVert_\infty}\right|^2}\leq 0$$ which with a bit of rearranging is the desired inequality.