# Proving a (seemingly simple) inequality [duplicate]

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.

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.
• $(x+y)^p-x^p-y^p$ is decreasing for $p\in(0,1)$. – cangrejo Feb 8 '19 at 9:09
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 $$y>0$$.