Inequality used to prove the lemma Brezis-Lieb

Question

How do I prove the inequality below? $$\left|\left|a+b\right|^p-\left|a\right|^p-\left|b\right|^p\right|\leq C~(\left|a\right|^{p-1}\left|b\right|+\left|a\right|\left|b\right|^{p-1} ), \ \forall \ a,b \in \mathbb R$$ where $C$ is a constant that depends on $p$ with $1< p < \infty$.

Answer 1

Hint :if "$|a|\le$d then $-d\le a\le d$ " $$1.{|{a+b}|^p - | a^p| - | b^p}|\le C \bigl(| a^{p-1}|| b |+ | a| | b^{p-1}\bigr)\color{red}{\iff_ 1}\color{green}{|a+b|^p\le |a|^{p-1}(c|b|+|a|)+|b|^{p-1}(c|a|+|b|)}$$$$2.{|{a+b}^p| - |a^p| - | b^p|}\ge -C \bigl(| a^{p-1}||b |+ |a| | b^{p-1}|\bigr)\color{red}{\iff_ 2}\color{green}{|{a+b}^p| \ge|a|^{p-1}(|b|-c|a|)+|b|^{p-1}(|a|-c|b|)}$$ clearly $\forall c\ge 0$$\color{red} {1,2}$ are true

Answer 2

Or by homogeneity, you just need to check that

$$\mathop {\sup }\limits_{|t| \leqslant 1} \frac{{\big| {|t + 1{|^p} - |t{|^p} - 1} \big|}}{{|t{|^{p - 1}} + |t|}} < \infty .$$