Inequality of $L^p$ type

Question

If $a\geq 1,$ $b\geq c\geq 1$ and $p>0$ then is it true that $$\frac{a+b}{\left\{\int_0^{2\pi}|e^{i\theta}+b|^pd\theta\right\}^{1/p}}\leq \frac{a+c}{\left\{\int_0^{2\pi}|e^{i\theta}+c|^pd\theta\right\}^{1/p}}? $$ In the maximum norm this is true, but I am not able to prove this case. Could anyone help me in this?

Answer 1

Let $p>0$. Raising both sides of the inequality to the power $-p$, we find that it is equivalent to $$ \int_0^{2\pi} \left(\frac{|e^{i\theta}+b|}{a+b}\right)^pd\theta \geq \int_0^{2\pi} \left(\frac{|e^{i\theta}+c|}{a+c}\right)^pd\theta. $$ Now to prove this inequality, we show that \begin{equation}\tag{1} \frac{|e^{i\theta}+b|}{a+b}\geq \frac{|e^{i\theta}+c|}{a+c} \end{equation} for any $\theta\in[0,2\pi]$ and $a\geq 1$, $b\geq c\geq 1$. This can be done by studying the function $$ f:x\mapsto \frac{|e^{i\theta}+x|}{a+x} $$ defined on $[1,+\infty[$. We find that $$ f'(x)\geq 0 \iff x(a-\cos\theta)+a\cos\theta-1 \geq 0, $$ which is shown to be always true since $\theta\mapsto x(a-\cos\theta)+a\cos\theta-1$ reaches its minimum either at $\theta=0$ or at $\theta=\pi$, and both corresponding values are positive. Thus both $f$ and $f^p$ are non-decreasing, $$ \left(\frac{|e^{i\theta}+b|}{a+b}\right)^p\geq \left(\frac{|e^{i\theta}+c|}{a+c}\right)^p $$ and the inequality is proved.

I believe that $(1)$ can be proven more simply with geometric considerations, but this works.