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I want to show that for $p \in (0,1)$,$(x +y)^p \leq x^p +y^p$. I thought of doing this:

Since $p \in (0,1)$, then $\frac{1}{p} \in (1,\infty)$. I can then raise both sides of the inequality to the $\frac{1}{p}$ power:$((x +y)^p)^{\frac{1}{p}} \leq (x^p +y^p)^{\frac{1}{p}} \leq x + y$, the last inequality would then follow from Jensen's inequality. It can be shown that $x^\frac{1}{p}$ is convex, so Jensen's would apply.

However, I don't think this is right, it is too easy (and I don't think proves anything). The difficulty here that I notice is that $p \in (0,1)$, so any argument using convexity won't work.

If anyone has any hints or suggestions, they are most welcomed.

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I suppose your $x$ and $y$ are non-negative. Show that $(x+y)^{p}-x^{p}-y^{p}$ is an decreasing function of $x$ (for fixed $y$) and it is $0$ when $x=0$. You cannot use convexity of $x^{1/p}$ to prove this.

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  • $\begingroup$ Makes sense, I was thinking about something similar yesterday using differentiation but second-guessed myself. Thank you. $\endgroup$ – Vegeta the Prince of Saiyans Feb 8 '19 at 9:00
  • $\begingroup$ It's the other way around, isn't it? $\endgroup$ – cangrejo Feb 8 '19 at 9:05
  • $\begingroup$ What is the other way around? $\endgroup$ – Vegeta the Prince of Saiyans Feb 8 '19 at 9:07
  • $\begingroup$ $(x+y)^p-x^p-y^p$ is decreasing for $p\in(0,1)$. $\endgroup$ – cangrejo Feb 8 '19 at 9:09
  • $\begingroup$ @broncoAbierto Yes. I meant decreasing. I have corrected the answer. Thanks. $\endgroup$ – Kavi Rama Murthy Feb 8 '19 at 9:11
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For any fixed (positive) $x$, the two functions $(x+y)^p$ and $x^p+y^p$ are equal at $y=0$, and their $y$-derivatives are $p(x+y)^{p-1}$ and $py^{p-1}$, respectively; it's easy to see that the second expression is always larger than the first expression (since $p<1$). Therefore the second function is always larger than the first function for $y>0$.

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