# How can we show that $\left|a+b+c\right|^p-2\left|a\right|^p\le C\left(\left|b\right|^p+\left|c\right|^p\right)$?

Let $$p\ge2$$. How can we show that $$\left|a+b+c\right|^p-2\left|a\right|^p\le C\left(\left|b\right|^p+\left|c\right|^p\right)\;\;\;\text{for all }a,b,c\in\mathbb R\tag1$$ for some $$C\ge0$$?

I'm only aware of similar inequalities like $$\left(\sum_{i=1}^nx_i\right)^2\le n\sum_{i=1}^nx_i^2\tag2$$ (obtained by the Cauchy-Schwarz inequality) and $$\left(\frac{\sum_{i=1}^nx_i}n\right)^p\le\frac{\sum_{i=1}^nx_i^p}n\tag3$$ (obtained by Jensen's inequality) for all $$x\in\mathbb R^n$$ and $$n\in\mathbb N$$.

• fix $b,c$. then differentiate w.r.t. $a$ to maximize LHS. Then you get an expression with just $b,c$. Fix $c$ and differentiate w.r.t. $b$, etc – mathworker21 Dec 22 '18 at 13:43

Let $$\lambda\in (0,\frac{1}{2})$$ be the constant chosen later and $$f(x) = |x|^p$$. For $$p>1$$, we know that $$f$$ is convex. Hence, by Jensen's inequality, we have $$\begin{eqnarray} f(a+b+c) &=& f\left((1-2\lambda)\frac{a}{1-2\lambda}+\lambda \frac{b}{\lambda}+\lambda \frac{c}{\lambda}\right)\\&\leq& (1-2\lambda)f(\frac{a}{1-2\lambda})+\lambda f( \frac{b}{\lambda})+\lambda f( \frac{b}{\lambda})\\ &=&(1-2\lambda)^{1-p}|a|^p + (\lambda^{1-p}|b|^p+\lambda^{1-p}|c|^p). \end{eqnarray}$$ If we choose $$\lambda$$ so that $$(1-2\lambda)^{p-1} =\frac{1}{2},$$ Then the condition is satisfied by $$C=\lambda^{1-p}.$$