Showing the inequality $|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)$

Solution 1:

• The map $x\mapsto x^p$ for $x\geq 0$ is convex, since its second derivative is $p(p-1)x^{p-2}\geq 0$.
• We have $$\left|\frac{a+b}2\right|^p\leq \left(\frac{|a|+|b|}2\right)^p\leq \frac{|a|^p+|b|^p}2.$$

Solution 2:

In fact, one advantage of the argument in john w.'s answer is that it can be extended further to prove $$(a+b)^n\leq p^na^n+q^nb^n,$$ where $\frac{1}{p}+\frac{1}{q}=1$. In particular, one can choose the coefficient of $a^n$ very close to $1$ by paying the price of a larger coefficient of $b^n$.

The proof is achieved by arguing either $(a+b)\leq pa$ or $(a+b)\leq qb$ holds. This is true since otherwise one would get a contradiction that $\left(\frac{1}{p}+\frac{1}{q}\right)(a+b)>a+b$.