# To show $(x+y)^p\leq x^p+y^p$, where $0\leq p\leq1, x>0,y>0$? [duplicate]

How to show that, $(x+y)^p\leq x^p+y^p$, where for $0\leq p\leq 1,x\geq 0, y\geq0?$ Any suggestion how to prove it?

Thanks in advance.

• This is wrong. Take $p = 0.5, x=1, y=1$. – Traklon Oct 24 '14 at 13:54
• @Kaushik, you should compute again - $LHS=4$ for $p=0.5,x=y=1$. – Galc127 Oct 24 '14 at 14:00
• @Kaushik he used $p=0.5$, not $p=2$. In which case we have $(1+1)^2 \leq 1^2 + 1^2$. – FH93 Oct 24 '14 at 14:00
• I think it should be $(x+y)^p\le x^p+y^p$ for $0\le p\le 1$. This is the Minkowski Inequality – Quang Hoang Oct 24 '14 at 14:03
• @Kaushik : The purpose of my comment was to indicate that maybe you wanted to type $p\geq 1$, not the opposite (or write $p$ instead of $\frac{1}{p}$). – Traklon Oct 24 '14 at 14:03

## 1 Answer

This should either be $(x+y)^p \le x^p + y^p$ for $0 < p \le 1$ or $(x+y)^{1/p} \le x^{1/p} + y^{1/p}$ for $p \ge 1$. We will use the former.

The common method is to prove it first for $y=1$. Define $\phi(x) = x^p + 1 - (x+1)^p$ for $x \ge 0$. Then $$\phi'(x) = px^{p-1} - p(x+1)^{p-1}.$$ Since $p-1 < 0$ you get $\phi'(x) > 0$ for all $x > 0$. Thus $\phi$ is increasing on $(0,\infty)$ so that $\phi(x) \ge \phi(0) = 0$ for all $x \ge 0$. That means $x^p + 1 - (x+1)^p \ge 0$ for all $x \ge 0$.

If $x,y > 0$ this implies $$\left(\frac xy + 1\right)^p \le 1 + \left( \frac xy \right)^p.$$ Multiply by $y^p$ to get the general inequality.

• Thanks to every one for your efforts. – Janak Oct 24 '14 at 14:08