An inequlaity involving an $L^p$ term

Question

For any positive real numbers $A, B $ and $p>0$ we have $$\int_0^{2\pi}|A+Be^{i\gamma}|^pd\gamma \geq 2\pi\max\{A^p, B^p\}.$$ Proof. Without loss of generality we can assume that $A\geq B> 0.$ Take $f(z)=(A+Bz)^{p/2}.$ From the inequality $$\int_0^{2\pi}|f(e^{i\gamma})|^2d\gamma\geq 2\pi|f(0)|^2$$ the above inequality follows. Any alternative interesting approach is welcome and appreciated.

Answer 1

Without losing the generality, assume $A \geq B > 0$ and set $r = \frac{B}{A} \in (0, 1]$. Then consider the function

$$ f(p) = \int_{0}^{2\pi} |1 + re^{i\theta}|^p \, \mathrm{d}\theta. $$

Since $p \mapsto |1 + re^{i\theta}|^p$ is convex for each choice of $r$ and $\theta$, it follows that $f(p)$ is also convex. Moreover,

$$ f'(0) = \operatorname{Re} \left[ \int_{0}^{2\pi} \log(1 + re^{i\theta}) \, \mathrm{d}\theta \right] = \operatorname{Re} \left[ \int_{|z|=r} \frac{\log(1 + z)}{iz} \, \mathrm{d}z \right] = 0. $$

Hence $f(0) = 2\pi$ is the global minimum of $f(p)$ and we conclude

$$ \int_{0}^{2\pi} |1 + re^{i\theta}|^p \, \mathrm{d}\theta = f(p) \geq f(0) = 2\pi, $$

which is equivalent to the desired inequality.