Thread: How 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

Originally Posted by Karima

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 .