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
There are a few fun ways to prove this.
First, consider the case .
Note then that
and
.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.
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 .