# $(x+y)^p\leq x^p+y^p$ for positive x, y and 0<p<1 [duplicate]

I need to use an inequality $(x+y)^p\leq x^p+y^p$ for positive $x$, $y$ and $0<p<1$ in a proof, but I'm not sure how to derive it and I wasn't able to find it anywhere. I think it might hold, because I put a number of queries with random rationals like (x+y)^(0.767809)<=x^(0.767809) + y^(0.767809) into Wolfram Alpha as well as $p=0.1$, ..., $0.9$ and it always said the inequalities were true for positive $x$ and $y$. I tried transforming it to use concavity/convexity, but got nowhere. Could somebody point me to a proof somewhere or give me a hint?

• Rewrite your inequality as a fraction which you must show to be greater than 1. Then treat it as a function of one variable by fixing the other one and take its derivative. You will find it will have a minimum in the range x and y positive. Evaluating the minimum, you will find that it is greater than one as long as p<1 which is given in the problem. I hope that helps. – Asier Calbet Aug 31 '14 at 11:59
• It's a variation on the beginning of the proof of Minkowski inequality, see here – Jean-Claude Arbaut Aug 31 '14 at 12:08

Take the function $f(x)= (x+y)^{p} -x^{p}-y^{p}$ then $$f'(x)= p\left[(x+y)^{p-1}-x^{p-1} \right]\leq 0$$ because $x\mapsto x^{p-1}$ is decreasing (since $p-1<0$) therefore : $$f(x) \leq f(0) = 0$$ Which is the desired result. And you can do the same thing if $p>1$ but the inequality would be reversed.
Divide both sides by $y^p$, and then set $a=\dfrac{x}{y}$. You will get $(a+1)^p\le a^p+1$, now you can derive $p$ which is more easy now.