Useful but not well known inequality

Question

At some point, I was looking in this forum for some useful inequality concerning absolute values. I found a very nice one, slightly simple to prove, but didn't find any formal proof of it, so I am using this question to write one here.

For every $x,y \in \mathbb{R}$ and any $p\geq 1$, the following inequality holds \begin{equation} 2^{1-p}|x-y|^p \leq |x|x|^{p-1}-y|y|^{p-1}|. \end{equation}

Context:

• I found first this inequality here: Difference of powers inequality , where they called it ''Mazur's inequality"
• In this other post Difference of powers inequality , someone cited these two places as a reference for this inequality, but they are only accesible in Chinese: https://www.zhihu.com/question/474094278/answer/2015826434 , https://www.doc88.com/p-096371580918.html (Page 135, (3rd Ed.))
• This other question Inequality for difference of powers , suggested some hints for the proof.

Answer 1

Note that since $p\geq 1$, the function $f: \mathbb{R}\rightarrow \mathbb{R}$ given by \begin{equation*} f(x)=|x|^p \end{equation*} is convex and continuous. In particular this implies that $f$ is mid-point convex (see Midpoint-Convex and Continuous Implies Convex), this is: for every $x,y \in \mathbb{R}$ we have \begin{equation}\label{midpoint} f\left( \frac{x+y}{2}\right)\leq \frac{f(x)+f(y)}{2}. \end{equation} We assume in this proof that $x\geq y$. This is not restrictive at all, since the proof in the case $y>x$ works analogous. We have then three different cases to consider.

• If $x\geq y \geq 0$: we write $\delta=x-y\geq0$ and then it is well known that \begin{equation*} \delta^p+y^p\leq (y+\delta)^p, \end{equation*} from where we get \begin{equation}\label{ineq1} (x-y)^p \leq x^p-y^p\leq 2^{p-1}(x^p-y^p), \end{equation} where the last inequality follows since $p\geq 1$. Now is just a matter of rewriting this and we obtain the statement.
• If $x\geq 0 \geq y$, then we can just write $\tilde{y}=-y>0$, and by the mid-point convexity we get \begin{equation*} \frac{|x-y|^p}{2^p}=\frac{|x+\tilde{y}|^p}{2^p}\leq \frac{|x|^p+|\tilde{y}|^p}{2}=\frac{x|x|^{p-1}-y|y|^{p-1}}{2}, \end{equation*} obtaining the result.
• The case $0\geq x\geq y$ is covered when $y\geq x\geq0$.