is it true that for all $p \geq 1, a,b \geq 0: (a+b)^{p} \leq 2^{p-1}(a^{p}+b^{p})$? If so, could someone give me a proof and an intuition of why this is true?
Thanks!
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This answer addresses your question, although the question there only asks about $(a+b)^p\le2^p\left(a^p+b^p\right)$. – robjohn Oct 03 '22 at 15:27
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The inequality can be written as $$ \left(\frac{a+b}2\right)^p\leqslant\frac 12a^p+\frac 12b^p $$ which is (since $p\geqslant 1$), nothing but convexity of the map $t\in [0,\infty)\mapsto t^p$.
If we do not know about convexity, it is still possible to show the inequality by assuming without loss of generality that $b=1$ and discuss the sign of the difference of the involved terms but looking at the derivative.
Davide Giraudo
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