# The complex numbers inequality $(|a+tb|^p-|a|^p)/t \leq |a+b|^p-|a|^p$

The following inequality is from the proof that the $L^p$ norm is Gâteaux differentiable for $1 < p<\infty$ (from "Analysis" by Lieb and Loss).

Let $a$, $b\in\mathbb{C}$ and $-1\leq t\leq 1$, $t\not=0.$ Then $$|a|^p-|a-b|^p\leq\frac{1}{t}(|a+tb|^p-|a|^p) \leq |a+b|^p-|a|^p.$$

I managed to prove the second inequality for positive $t$ by writing $a+tb=(1-t)a+t(a+b)$and using the convexity of $\cdot^p$. From this the first inequality follows for negative $t$ by substituting $-b$ for $b$. The same trick would finish the proof, if I could prove either the second inequality for negative $t$ or the first inequality for positive $t$.

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The same basic idea works for negative values of $t$:

What you want to prove is equivalent to

\begin{equation*} |a + tb|^p - |a|^p \geq t|a + b|^p - t|a|^p. \end{equation*}

This in turn is the same as

\begin{equation*} (1 - t)|a|^p \leq |a + tb|^p + (-t)|a + b|^p, \end{equation*}

which in turn is equivalent to

\begin{equation*} |a|^p \leq {1 \over 1 - t} |a + tb|^p + {-t \over 1 - t} |a + b|^p. \end{equation*}

Note that

\begin{equation*} a = {1 \over 1 - t} (a + tb) + {-t \over 1 - t} (a + b), \end{equation*}

so you can use convexity as you used before.

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