I would need a hint to prove the following statement :
For every $\epsilon>0$ there exists $C_{\epsilon}>0$ such that for every couple $a,b\ge 0$ we have $(a+b)^p\leq(1+\epsilon)a^p+C_{\epsilon}b^p$
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edamondo
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Also: https://math.stackexchange.com/q/2853822/42969 – Martin R Jun 03 '21 at 11:42
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@MartinR, thanks, Iam not sure I understood why "For $x>x_\varepsilon$, it is evident as well with $C_\varepsilon = (1+x_\varepsilon^{-1})^p$." in the first link – edamondo Jun 03 '21 at 23:17
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See https://math.stackexchange.com/questions/2853822/proof-of-the-inequality-abp-ap-leq-epsilon-ap-c-epsilon-b for a more detailed solution. – Mike Earnest Jun 07 '21 at 16:16
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@edamondo: This may be late, but I wrote an answer to this problem here based on Jensen's inequality. – Mittens Jan 27 '22 at 22:24
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@OliverDiaz, thanks! – edamondo Jan 27 '22 at 23:19
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Some thoughts: if either $a = 0$ or $b=0$ the result is trivial.
If $b \not = 0$ you can divide everything by $b$ and write $t = \dfrac ab$ to arrive at $$(t+1)^p \le (1+\epsilon)t^p + C_\epsilon.$$
What you need to show is that, for every $\epsilon > 0$, the function $$\phi_\epsilon(t) = (t+1)^p - (1+\epsilon)t^p,\quad t \ge 0,$$ is bounded above. Try the first derivative test.
Umberto P.
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