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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- January 24th 2013, 11:28 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 - January 24th 2013, 03:39 PMabenderRe: How can this inequality be proved?
There are a few fun ways to prove this.

.

First, consider the case

Note then that

.Since

,it follows that

,and therefore,

when .

.**Next, assume, Without Loss of Generality, that**

Since , we have an expansion of the form

, .

*I'm a bit tired, so perhaps you can finish this or I will later on.* - January 24th 2013, 04:49 PMjohngRe: How can this inequality be proved?
Let me rephrase your inequality so what follows is more clear: where and .

If , since -- the last inequality since the function is increasing.

So suppose . I can suppose a<b. (The problem is symmetric in a and b.) Consider the function on . Now --

since is decreasing. Since , the function f can be extended to a continuous function on where and otherwise. Now is decreasing on -- the proof that a function on , which has a negative derivative in the open interval , uses the mean value theorem on ; remember that differentiability is required only on the open interval. So for any for any or

Finally, and easily then .