How can we show $\left\|f-g\right\|_{L^p(\mu)}^p\le2^{p-2}\left\|f-g\right\|_{L^2(\mu)}^2$? if $p>2$ and $|f|,|g|\le1$? Let $(\Omega,\mathcal A,\mu)$ be a finite measure space, $p>2$ and $f,g:\Omega\to\mathbb R$ with $|f|,|g|\le1$.

How can we show that $\left\|f-g\right\|_{L^p(\mu)}^p\le2^{p-2}\left\|f-g\right\|_{L^2(\mu)}^2$?

The trick must obviously be writing $\left\|f-g\right\|_{L^p(\mu)}^p=2^p\left\|\frac{f-g}2\right\|_{L^p(\mu)}^p$ and noting that $|f-g|\le2$, but how do we need to proceed?

I believe it's simply $$\int|f-g|^p = \int|f-g|^2\cdot|f-g|^{p-2}\le 2^{p-2}\int|f-g|^2.$$
