# Prove: $\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$

Prove: $$\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right) \geq 2x^2$$ for $\: -\frac{\pi}{2} < x < \frac{\pi}{2}$

I have thought of three things so far that can be useful.

The Cauchy–Schwarz inequality $$\int_Efg\,\mathrm{d}x\le\left(\int_Ef^2\,\mathrm{d}x\right)^{1/2}\left(\int_Eg^2\,\mathrm{d}x\right)^{1/2}$$

Both sides of the inequality are even functions so its enough to consider $x \in [0, \frac{\pi}{2})$.

Maybe I can use that $\int_0^x \cos t \:dt = \sin x$.

I would appreciate a hint or a few on how to continue.

edit: The inequality is supposed to be proved using integration methods.

• Perhaps this will help. Mar 8, 2016 at 23:47
• @goblin It is solved already. Mar 8, 2016 at 23:50

$$\int_0^x \cos t \:dt = \sin x$$

$$\int_0^x \frac{1}{\cos t} \:dt = \frac{1}{2}\ln \left(\frac{1 + \sin x}{1 - \sin x}\right)$$

Let

$$f(t) = \sqrt{\cos t}$$

$$g(t) = \frac{1}{\sqrt{\cos t}}$$

Cauchy–Schwarz

$$\left(\int_0^x f(t)g(t) \:dt\right)^2 \leq \int_0^x f^2(t) \:dt \int_0^x g^2(t) \:dt$$

proves the result.

• +1. Wonderful proof.
– Hans
May 14, 2022 at 6:52

Without loss of generality, let $0\le x < \frac{\pi}{2}$. Define $$f(x)=\sin x \ln \left(\frac{1 + \sin x}{1 - \sin x}\right).$$ Then $f(0)=0,f'(0)=0$ and \begin{eqnarray} f''(x)=\frac{1}{8\cos^2x}(4 \cos (2 x)-\sin x \ln (\frac{1+\sin x}{1-\sin x})-\sin (3 x) \ln(\frac{1+\sin x}{1-\sin x})+12). \end{eqnarray} We want to show that $f''(x)\ge 4$. Note $$f''(x)-4=-\frac{1}{2} \tan x \sec x \left(-4 \sin (x)+\ln \left(\frac{1+\sin x}{1-\sin x}\right)+\cos (2 x) \ln \left(\frac{1+\sin x}{1-\sin x}\right)\right).$$ Define $$g(x)= 4 \sin x-\ln \left(\frac{1+\sin x}{1-\sin x}\right)-\cos (2 x) \ln \left(\frac{1+\sin x}{1-\sin x}\right).$$ Now we show that $g(x)\ge0$ for $0\le x<\pi/2$. In fact, easy calculation shows $$g'(x)=2\sin(2x)\ln \left(\frac{1+\sin x}{1-\sin x}\right)\ge0$$ for $0\le x<\pi/2$ and hence $g(x)$ is increasing. So $g(x)\ge 0$ and hence $$f(x)=f(0)+f'(0)+\frac{1}{2}f''(c)x^2=\frac{1}{2}f''(c)x^2\ge 2x^2$$ for some $c\in(0,\pi/2)$.

For $x^2\lt1$, \begin{align} x\log\left(\frac{1+x}{1-x}\right) &=\sum_{k=0}^\infty\frac{2x^{2k+2}}{2k+1}\\ &\le\sum_{k=0}^\infty2x^{2k+2}\\ &=\frac{2x^2}{1-x^2}\tag{1} \end{align} If $$f(x)=\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\tag{2}$$ then $$f'(x)=\cos(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\tan(x)\tag{3}$$ and applying $(1)$ \begin{align} f''(x) &=2-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)+2\sec^2(x)\\ &=4+2\tan^2(x)-\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\\[4pt] &\ge4+2\tan^2(x)-2\tan^2(x)\\[12pt] &=4\tag{4} \end{align}
Since $f(0)=f'(0)=0$ and $f''(x)\ge4$, we get that $f(x)\ge2x^2$. Therefore, applying $(1)$ again, we get $$2x^2\le\sin(x)\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\le2\tan^2(x)\tag{5}$$

Hint $$\ln\left(\frac{1+\sin x}{1-\sin x}\right)\approx 2\left(\sin x+\frac{\sin^3 (x)}{3}+...\right)$$ or use $\ln(1+x)=x-x^2/2+x^3/3...$ and $\ln(1-x)=-(x+x^2/2+x^3/3...)$ as $|x|<1$