If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.

It was on Wikipedia:

http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities

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- Jan 24th 2013, 10:28 AMKarimaHow can this inequality be proved?
If x, y > 0 and 0 < p < 1, then (x+y)^p < x^p+y^p.

It was on Wikipedia:

http://en.wikipedia.org/wiki/Inequality_(mathematics)#Power_inequalities - Jan 24th 2013, 02:39 PMabenderRe: How can this inequality be proved?
There are a few fun ways to prove this.

$\displaystyle \, x=y$.

First, consider the case

Note then that

$\displaystyle (x+y)^p = (x+x)^p = (2x)^p = 2^p(x^p)$and

$\displaystyle x^p + y^p = x^p + x^p = 2(x^p)$.Since

$\displaystyle 0<p<1$,it follows that

$\displaystyle 2^p<2$,and therefore,

$\displaystyle (x+y)^p < x^p+y^p $ when $\displaystyle x=y$.

$\displaystyle \, x>y$.**Next, assume, Without Loss of Generality, that**

$\displaystyle (x+y)^2 = \left(x \left(1 + \tfrac{y}{x} \right)\right)^p = x^p \left(1+\tfrac{y}{x}\right)^p $

Since $\displaystyle \tfrac{y}{x}<1$, we have an expansion of the form

$\displaystyle (1+r)^p$, $\displaystyle 0<r<1$.

*I'm a bit tired, so perhaps you can finish this or I will later on.* - Jan 24th 2013, 03:49 PMjohngRe: How can this inequality be proved?
Let me rephrase your inequality so what follows is more clear: $\displaystyle (a+b)^p<a^p+b^p$ where $\displaystyle a,b>0$ and $\displaystyle 0<p<1$.

If $\displaystyle a=b$, $\displaystyle 2^pa^p<2a^p$ since $\displaystyle 2^p<2$ -- the last inequality since the function $\displaystyle x^p$ is increasing.

So suppose $\displaystyle a\ne b$. I can suppose a<b. (The problem is symmetric in a and b.) Consider the function $\displaystyle f(x)=(x+1)^p-x^p$ on $\displaystyle (0,\infty)$. Now $\displaystyle f^{\prime}(x)<0$ --

$\displaystyle (x+1)^{p-1}-x^{p-1}<0$ since $\displaystyle x^{p-1}$ is decreasing. Since $\displaystyle \lim_{x\rightarrow 0^+}f(x)=1$, the function f can be extended to a continuous function $\displaystyle g$ on $\displaystyle [0,\infty)$ where $\displaystyle g(0)=1$ and $\displaystyle g(x)=f(x)$ otherwise. Now $\displaystyle g$ is decreasing on $\displaystyle [0,\infty)$ -- the proof that a function on $\displaystyle [0,\infty)$, which has a negative derivative in the open interval $\displaystyle (0,\infty)$, uses the mean value theorem on $\displaystyle [0,\infty)$; remember that differentiability is required only on the open interval. So for any $\displaystyle x>0, 1=g(0)>g(x)=(x+1)^p-x^p$ for any $\displaystyle x>0$ or $\displaystyle (x+1)^p<x^p+1$

Finally, $\displaystyle (a/b+1)^p<(a/b)^p+1$ and easily then $\displaystyle (a+b)^p<a^p+b^p$.