To prove the integral inequality $\int_\overline{\theta}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}\gt\pi$

Question

The following inequality comes up in connection with motion in a dipole field. One has to show that $$\int_\overline{\theta}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}\gt\pi$$ where $$\overline{\theta}= \begin{cases}0, & 0\lt\lambda \lt1\\ \arccos(\frac{1}{\lambda}), & \lambda \gt 1\end{cases}.$$ The case $0\lt\lambda\lt 1$ is easily verified. For then $$\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}+\int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}$$ $$\implies\int_{0}^{\pi}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \cos{\theta}}}+\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\lambda \sin{\theta}}}\gt2\int_{0}^{\pi/2}d\theta=\pi.$$ However I am finding it difficult to prove the second case. Any pointers would be helpful.

Thanks.

NOTE: The inequality does not hold for all values of $\lambda > 1$, as has been noted in the following responses.

Answer 1

If I divide by $\lambda$ the inequality becomes: $$\int_{\underline{\theta}}^{\pi}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\geq \pi \sqrt{\lambda}$$

Now $$\int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\leq \int_{\pi/2}^{\pi}\frac{d\theta}{\sqrt{-\cos(\theta)}}d\theta=\int_0^{\pi/2}\frac{d\theta}{\sqrt{\sin(\theta)}}<+\infty$$ and as if $\theta \in [ \underline{\theta}, \pi/2]$, we have $\cos(\theta)-\cos(\underline{\theta})=(\theta-\underline{\theta})(-\sin(c))$ for some $c\in [ \underline{\theta}, \pi/2]$, we get $|\cos(\theta)-\cos(\underline{\theta})|\geq(\theta-\underline{\theta})\sin(\underline{\theta}) $ Hence $$\int_{\underline{\theta}}^{\pi/2}\frac{d\theta}{\sqrt{\cos(\underline{\theta})-\cos(\theta)}}\leq\frac{1}{\sqrt{\sin(\underline{\theta})}} \int_{\underline{\theta}}^{\pi/2}\frac{d\theta}{\sqrt{\theta-\underline{\theta}}}= 2\frac{\sqrt{\pi/2-\underline{\theta}}}{\sqrt{\sin(\underline{\theta})}} $$

Now as $\displaystyle \sin(\underline{\theta})=\sqrt{1-\frac{1}{\lambda^2}}$, we have $\displaystyle 2\frac{\sqrt{\pi/2-\underline{\theta}}}{\sqrt{\sin(\underline{\theta})}}$ bounded as $\lambda\to +\infty$, so if I am not wrong, there is a problem.

Answer 2

Your inequality is wrong in general (indicated below), and your proof for $\lambda<1$ has a small mistake. Let us start with the latter.

The problem with your proof

You divide the integral in the intervals $(0,\pi/2)$ and $(\pi/2,\pi)$ and that is fine. But since $\cos \theta<0$ in the interval $(\pi/2,\pi)$ it is impossible that you get as an integrand $1/\sqrt{1-\lambda\sin \theta}$ after your change of variables. What you get if you just shift $\theta\mapsto \theta-\pi/2$ is $1/\sqrt{1+\lambda\sin\theta}$. But this is less than $1$ on the interval. Instead, let $\theta\mapsto \pi-\theta$. Then, you will end up with $$ \int_0^{\pi/2}\frac{1}{\sqrt{1-\lambda\cos\theta}}+\frac{1}{\sqrt{1+\lambda\cos\theta}}\,d\theta. $$ Then, according to the inequality $a+b\geq 2\sqrt{ab}$, you find that your integrand is bounded below by $$ \frac{2}{\sqrt[4]{1-\lambda^2\cos^2\theta}} $$ which is greater than $2$ on the interval. Integrating you get a lower bound of $\pi$.

Proof that the inequality is not true for all $\lambda>1$

Here, as suggested by @Ian, it is sufficient to consider $\lambda=2$, and hence show that the integral $$ \int_{\pi/3}^\pi\frac{1}{\sqrt{1-2\cos\theta}}\,d\theta<\pi. $$ In the interval $(\pi/3,\pi)$ it holds that $$ 1-2\cos\theta>\frac{9}{2\pi}(\theta-\pi/3). $$ Thus $$ \int_{\pi/3}^\pi\frac{1}{\sqrt{1-2\cos\theta}}\,d\theta< \int_{\pi/3}^\pi\frac{1}{\sqrt{\frac{9}{2\pi}(\theta-\pi/3)}}\,d\theta=\frac{4\pi}{3\sqrt{3}}<\pi. $$