Inequality of $L^p$ type 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?
 A: 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. 
