Let $1 \le p < \infty$, and let $a, b \ge 0$. Prove $(a + b)^p \le (1 + \epsilon) a^p + C_\epsilon b^p$.

Asked May 25 '20 at 16:39

Active May 25 '20 at 16:53

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Let $1 \le p < \infty$, and let $a, b \ge 0$. I would like to prove that for any $\epsilon > 0$, there exists a $C_\epsilon > 0$ depending on $\epsilon$ such that $$ (a + b)^p \le (1 + \epsilon) a^p + C_\epsilon b^p $$

I could not find a reference for this inequality or a name, so if anyone knows where I could find a proof that would be helpful.

edited May 25 '20 at 16:53

asked May 25 '20 at 16:39

Austin David Brown

So $C_{\epsilon}$ does not depend on $a,b$ and $p$? – N.Quy May 25 '20 at 17:06
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This boils down to $(1+x)^p\leq (1+\epsilon)+Cx^p$, which is elementary because univariate – Bananach May 25 '20 at 17:11
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This was asked and answered here https://math.stackexchange.com/questions/2853822/proof-of-the-inequality-abp-ap-leq-epsilon-ap-c-epsilon-b?rq=1 – Mittens May 25 '20 at 17:17
Thanks for the help! – Austin David Brown May 25 '20 at 17:31

Let $1 \le p < \infty$, and let $a, b \ge 0$. Prove $(a + b)^p \le (1 + \epsilon) a^p + C_\epsilon b^p$.

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