# Showing $\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p$

For $a,b \in \mathbb R$, $p\geq2$ I try to show $$\left|\frac{a+b}{2}\right|^p+\left|\frac{a-b}{2}\right|^p\leq\frac{1}{2}|a|^p+\frac{1}{2}|b|^p.$$

Is this a popular inequality (At least I could not find it in the list of popular inequalities from wikipedia)? It seems to be related to convexity but I did not succeed to show it. A related inequality seems to be for $p \geq 1,a,b\geq0$

$$\left(\frac{a+b}{2}\right)^p\leq \frac{1}{2}a^p+\frac{1}{2}b^p,$$ which directly follows from the convexity of $x^p$ for positive numbers.

• It's a special case of Clarkson's inequalities (when your measure space is a single point with counting measure). Dec 31, 2011 at 5:08

We have for all $x_1,x_2\geq 0$: $x_1^p+x_2^p\leqslant (x_1^2+x_2^2)^{p/2}$. Indeed, it suffice to show it when $x_2=1$, otherwise apply it to $\frac{x_1}{x_2}$. $f(t):=(t^2+1)^{p/2}-t^p-1$ is non-negative, since its derivative is $p(t^2+1)^{p/2-1}t-pt^{p-1}\geqslant 0$ and $f(0)=0$. Applying it to $x_1=\frac{a+b}2$ and $x_2=\frac{a-b}2$, we get \begin{align*} \left|\frac{a+b}2\right|^p+\left|\frac{a-b}2\right|^p&\leqslant \left(\left|\frac{a+b}2\right|^2+\left|\frac{a-b}2\right|^2\right)^{p/2}\\\ &=\left(\frac {2a^2+2b^2}4\right)^{p/2}\\\ &=\left(\frac {a^2}2+\frac{b^2}2\right)^{p/2}\\\ &\leqslant \frac 12|a|^p+\frac 12|b|^p, \end{align*} since the map $t\mapsto |t|^{p/2}$ is convex ($p\geqslant 2$).
• Why is it sufficient to show it for $x_2=1$? Dec 31, 2011 at 0:10
• It's clear if $x_2=0$. Otherwise, you apply it to $\frac{x_1}{x_2}$. Dec 31, 2011 at 0:12
• Does the last step even follow from the convexity of $t\mapsto |t|^{p/2}$? I don't see it :/ Dec 31, 2011 at 14:42