I want to prove the inequality $$ |x-y|^p \le \frac{p}{2}\big|x-y\big|\;\big(x^{p-1}+y^{p-1}\big) $$ if $p \ge 1$. For the case $p$ is an integer, it's easy to do, but I have no idea when $p$ is not an integer! Thank you for the answer.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
0
Suppose $y>x\ge 0$. Then we have $$(y-x)^p\le y^p-x^p=\int_x^y pz^{p-1}dz.$$ For any $p\ge 1$, the right hand side is bounded by $$p(y-x)y^{p-1}.$$ For $p\ge 2$, the function $z\mapsto z^{p-1}$ is convex, and so the integral is bounded by its trapezoidal approximation. This leads to the bound $$p(y-x){x^{p-1}+y^{p-1}\over 2}.$$
answered Jan 26, 2014 at 15:57
$\begingroup$
$\endgroup$
1
It is false for 1 < p < 2. To see this, simply set x=0.
answered Jan 26, 2014 at 13:05
-
$\begingroup$ sorry, I modified the question, I try the case that p is an integer and thought that the right can be improved, what if the right is multiplied by 2? $\endgroup$– Chen JieCommented Jan 26, 2014 at 13:18