Inequalities with absolute values in $\mathbb{R}$

Question

Let $a, b, p$ be real numbers, with $p \geq 1$. Show that $ |ta+(1-t)b|^p \leq t|a|^p + (1-t) |b|^p$ for $0 \leq t \leq 1$. In particular $|a+b|^p \leq 2^{p-1} (|a|^p + |b|^p)$.

Answer 1

The main point is that the function $f(x)=|x|^p$ is convex (the second derivative is positive for all $x\neq 0$). Then the conclusion follows by the inequality of convex functions (https://en.wikipedia.org/wiki/Convex_function): For all ${\displaystyle 0\leq t\leq 1}$ and all ${\displaystyle x_{1},x_{2}\in \mathbb R}$, ${\displaystyle f\left(tx_{1}+(1-t)x_{2}\right)\leq tf\left(x_{1}\right)+(1-t)f\left(x_{2}\right)}$.

You can also see a direct proof of the inequality here Elementary proof that $|x|^p$ is convex.