Inner (dot) Product Inequality: Proving Nonnegativity

[babyrudin]

Homework Statement

For $$x,y \in R^n$$, their inner ("dot") product is given by

$$<x,y>=\sum_{i=1}^n x_i y_i.$$

Also, we write

$$<x,x>=\|x\|^2.$$

Homework Equations

Fix p>1. Show that for all $$x,y \in R^n$$ we have

$$< \|x\|^{p-2}x -\|y\|^{p-2}y, x-y> \geq 0$$

The Attempt at a Solution

Expanding the left-hand side, we can write

$$<\|x\|^{p-2}x,x> -<\|y\|^{p-2}y,x>-<\|x\|^{p-2}x,y>+<\|y\|^{p-2}y,y>$$

which further simplifies to

$$\|x\|^p +\|y\|^p -(\|x\|^{p-2} +\|y\|^{p-2})<x,y>.$$

Then I'm stuck. How do I show that the above is nonnegative?

 

[MathematicalPhysicist]

how about using cauchy-schwartz inequality?
something like this
<x,y><=sqrt(<x,x><y,y>)=||x||*||y||
then you get that what you wrote is greater than:
||x||^p+||y||^p-(||x||^p-2+||y||^p-2)(||x||*||y||)=
x^(p-1)(x-y)+y^(p-1)(y-x)=(x^p-1-y^p-1)(x-y)
now if ||x||>=||y|| then ||x||^p-1>=||y||^p-1
I leave you to check this proposition.

 

[babyrudin]

I've got it now, many many thanks!