If$\,\,$ $0<p<\infty$, put$\,\,$ $\gamma_{p}=\max(1,2^{p-1})$, and show that $$|\alpha-\beta|^p \leq \gamma_{p}(|\alpha|^p + |\beta|^p)$$ for arbitrary complex numbers $\alpha$ and $\beta$.
What I know is that $$|\alpha-\beta|^p \leq|2\max(|\alpha|,|\beta|)|^p=2^p\max(|\alpha|,|\beta|)^p=2^p\max(|\alpha|^p,|\beta|^p)\leq2^p(|\alpha|^p + |\beta|^p)$$
Can you help me please?
I hope I didnt miss something, but this should work,
For $p=1$ the inequality trivialy holds and notice that $\forall p>0$ we have,
$$|a-b|^p\leq|a+b|^p$$
We have two cases to consider,
(1) $p>1$:
Note that $h(x)=x^p$ is convex and monotone for $p>1$.
Hence,
$$|\frac{1}{2}a+\frac{1}{2}b|^p\leq|\frac{1}{2}|a|+\frac{1}{2}|b||^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$$ $$\Leftrightarrow|a-b|^p\leq|a+b|^p\leq2^{p-1}(|a|^p+|b|^p)$$
(2) $0<p<1$
In this case $h(x)=x^p$ is just monotone but $h(x)=x^p\leq x $.
Hence,
$$|a-b|^p\leq|a+b|^p\leq||a|+|b||^p\leq||a|^p+|b|^p|^p\leq(|a|^p+|b|^p)$$
What finally showes what was to prove.
I hoped i didnt miss something, do not hesitate correcting me...