# If $\alpha$ and $\beta$ are complex numbers, then $|\alpha-\beta|^p\leq(|\alpha|^p+|\beta|^p)$?

I have no idea how to show this in the case that $0<p<1$. I can show that $|\alpha-\beta|^p\leq 2^{p-1}(|\alpha|^p+|\beta|^p)$ for $p\geq1$ by using the convexity of the function $x^p$, but am lost at how to solve for this specific case. Any guidance would be much appreciated. Thanks!

$$|\alpha - \beta|^p \leq (|\alpha| + |\beta|)^p$$
We want to show $$(|\alpha| + |\beta|)^p \leq |\alpha|^p + |\beta|^p$$
which is equivalent to $$1 \leq (\dfrac{|\alpha|}{|\alpha| + |\beta|})^p + (\dfrac{|\beta|}{|\alpha| + |\beta|})^p$$
when $0< \lambda<1$ and $0 < p<1$, we have $\lambda^p > \lambda$, so $$(\dfrac{|\alpha|}{|\alpha| + |\beta|})^p \geq \dfrac{|\alpha|}{|\alpha| + |\beta|}$$ and $$(\dfrac{|\beta|}{|\alpha| + |\beta|})^p \geq \dfrac{|\beta|}{|\alpha| + |\beta|}$$