I am trying to proove that $x=0$ is the minimum point of the function
$$f(x) = \cos^2(x)\left( \csc^2(x_+) + \csc^2(x_-) \right)$$
in the interval $0 \leq x < \pi/2$ where $x_\pm = \frac\pi{2n} \pm \frac x n$,
and $n \in \mathbb{N}$ such that $n \geq 3$.
I have unsuccessfully tried the following approaches.
- Attempted to show that $f(x) – f(0) \geq 0$, i.e.
$$\cos^2(x)\left(
\csc^2\left(\frac\pi{2n} – \frac x n\right) +
\csc^2\left(\frac\pi{2n} + \frac x n\right)
\right) – 2\csc^2\left(\frac\pi{2n}\right) \geq 0;$$ - Computed $f'(x) = \frac\partial{\partial x} f(x)$ and tried to show that $f'(x) \geq 0$ in the interval, where
\begin{align}
\frac\partial{\partial x} f(x) = 2\cos^2(x) \left(
\frac{1}{n} \cot(x_-)\csc^2(x_-) –
\frac{1}{n} \cot(x_+)\csc^2(x_+) –
\tan(x)\left(\csc^2(x_-) + \csc^2(x_+)\right)
\right);
\end{align} - Used Taylor Series Expansion in $f(x)$ attempting to find a tight lower bound for $f(x)$.
Any help or hint would be appreciated.
Best Answer
Proof:
We need to prove that $f(x) \ge f(0) = 2\csc^2\frac{\pi}{2n}$ for all $0 \le x < \frac{\pi}{2}$.
If $n = 3$, we have $$f(x) = - 16\left(\cos^2 \frac{x}{3} - \frac34\right)^2 + 9 \ge 8 = 2\csc^2 \frac{\pi}{6}$$ where we have used $\frac{3}{4} \le \cos^2 \frac{x}{3} \le 1$.
In the following, assume that $n \ge 4$.
We have $$f(x) = \cos^2 x \cdot \frac{2\cos\frac{\pi}{n}\sin^2 \frac{x}{n} + 2\sin^2 \frac{\pi}{2n}}{(\sin^2 \frac{\pi}{2n} - \sin^2 \frac{x}{n})^2}.$$
Fact 1: $\sin \frac{x}{n} \ge \frac{2}{\pi} x \sin \frac{\pi}{2n}$.
(Hint: Take derivative.)
Fact 2: $\cos^2 x \cdot \frac{4x^2\pi^2\cos \frac{\pi}{4} + \pi^4}{\left(\pi^2 - 4x^2 \right)^2} \ge 1$.
(The proof is given at the end.)
By using Fact 1, we have \begin{align*} f(x) &\ge \cos^2 x \cdot \frac{2\cos\frac{\pi}{n}\cdot (\frac{2}{\pi} x \sin \frac{\pi}{2n})^2 + 2\sin^2 \frac{\pi}{2n}}{\left[\sin^2 \frac{\pi}{2n} - (\frac{2}{\pi} x \sin \frac{\pi}{2n} )^2\right]^2}\\[5pt] &= 2\csc^2\frac{\pi}{2n} \cdot \cos^2 x \cdot \frac{\frac{4x^2}{\pi^2}\cos\frac{\pi}{n} + 1}{\left(1 - \frac{4x^2}{\pi^2} \right)^2}\\[5pt] &\ge 2\csc^2\frac{\pi}{2n} \cdot \cos^2 x \cdot \frac{\frac{4x^2}{\pi^2}\cos \frac{\pi}{4} + 1}{\left(1 - \frac{4x^2}{\pi^2} \right)^2}\\[5pt] &\ge 2\csc^2\frac{\pi}{2n} \tag{1} \end{align*} where we have used Fact 2 to obtain (1).
We are done.
Proof of Fact 2:
Fact 3: $\cos x \ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1 \ge 0$ for all $0 \le x < \pi/2$.
(The proof is given at the end.)
By using Fact 3, it suffices to prove that $$\left(\frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1\right)^2\cdot \frac{4x^2\pi^2\cos \frac{\pi}{4} + \pi^4}{\left(\pi^2 - 4x^2 \right)^2} \ge 1$$ or $$\frac{x^2}{4\pi^6}(Ax^4 + Bx^2 + C) \ge 0$$ where \begin{align*} A &= 2\,\sqrt{2}\, {\pi }^{4}-32\,\sqrt {2}\,{\pi }^{2}+128\,\sqrt {2}, \\ B &= {\pi }^{6}-8\,\sqrt{2}\,{\pi }^{4}-16\,{\pi }^{4}+64\,\sqrt{2}\,{\pi }^{ 2}+64\,{\pi }^{2} ,\\ C &= -4\,{\pi }^{6}+8\,\sqrt {2}\,{\pi }^{4}+32\,{\pi }^{4}. \end{align*} It is easy to prove that $Ax^4 + Bx^2 + C \ge 0$ for all $0 \le x < \pi/2$.
We are done.
Proof of Fact 3:
We split into two cases:
It is easy to prove that $$\cos x \ge 1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 - \frac{1}{720}x^6.$$ It suffices to prove that $$1 - \frac{1}{2}x^2 + \frac{1}{24}x^4 - \frac{1}{720}x^6 \ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1$$ which is true (easy).
It is easy to prove that $$\cos x \ge - \left(x - \frac{\pi}{2}\right) + \frac{1}{6}\left(x - \frac{\pi}{2}\right)^3.$$ It suffices to prove that $$- \left(x - \frac{\pi}{2}\right) + \frac{1}{6}\left(x - \frac{\pi}{2}\right)^3 \ge \frac{2\pi^2 - 16}{\pi^4}x^4 - \frac{1}{2}x^2 + 1$$ which is true (easy).
We are done.