I am currently reading Lieb-Loss' book on Analysis. In the proof of Theorem 1.9 (Brézis-Lieb Lemma), whose statement is not relevant here, they use the following statement:
Let $p \in (0,\infty)$. For any $\varepsilon>0$, there is a constant $C_\varepsilon$ so that \begin{equation}\lvert\lvert a+b\rvert^p - \lvert b\rvert^p\rvert \leq \varepsilon \lvert b\rvert^p + C_\varepsilon \lvert a \rvert^p \ \ \ \ \ (1)\end{equation} for all complex numbers $a$ and $b$.
To prove the existence of such constant $C_\varepsilon$ in the case $p>1$, they use the inequality $$\lvert a + b\rvert^p \leq (1-\lambda)^{1-p}\lvert a\rvert^p + \lambda^{1-p}\lvert b \rvert^p \ \text{for all} \ 0<\lambda < 1,$$ which follows from the convexity of the map $t\mapsto \lvert t \rvert^p$. They claim that, to obtain (1), it suffices to take $\lambda = (1+\varepsilon)^{-1/(p-1)}$. I see that this choice yields $$\lvert a + b \rvert^p - \lvert b\rvert^p\leq \varepsilon \lvert b \rvert^p+\left(1-\frac{1}{(1+\varepsilon)^{1/(p-1)}}\right)^{1-p}\lvert a\rvert^p. \ \ \ (2)$$ I cannot see, however, how we obtain (2) with $\lvert b \rvert^p - \lvert a + b \rvert^p$ on the left hand side instead of $\lvert a + b \rvert^p - \lvert b \lvert^p$. Can anyone explain how to finish this?