p-norm Inequality

Question

Let $x,y \in \mathbb{R}$, and $0<p<\infty$. Prove $$|x-y|^p\leq (1+2^p)(|x|^p+|y|^p) $$

The case $0<p\leq1$ is obvious, as it follows from the properties of the P-norm, where $P:=1/p$, $$|x-y|^p=|x-y|^{1/P}\leq |x|^{1/P} + |y|^{1/P} = |x|^{p} + |y|^{p} \leq (1+2^p)(|x|^p+|y|^p)$$. But I'm stumped for the case $1<p<\infty$. Could someone point me in the right direction? Hope I'm not missing something obvious.

Answer 1

Thanks to @Potato's suggestion, we have $$\left|x-y\right|^p \leq \left( |x|+|y|\right)^p \leq 2^p\left(\frac{1}{2}\left|x\right|^p + \frac{1}{2}\left|2y\right|^p \right)= 2^{p-1}(|x|^p+|y|^p) \leq (1+2^p)(|x|^p+|y|^p)$$ for the case $1<p<\infty$.