I have the following problem:

x is a real number that is not equal to 0.

Prove that:
x^2+ (1⁄x^2) ≥ 2

I know the equation is right, but I can't seem to figure out how to prove it.

I have the following problem:

x is a real number that is not equal to 0.

Prove that:
x^2+ (1⁄x^2) ≥ 2
similar: prove that math]x^2+ (1⁄x^2) -2 ≥ 0[/math]

I know the equation is right, but I can't seem to figure out how to prove it.

This would be my approach.
First, find d/dx and d^2/dx^2

set d/dx=0 (this will get you either the highest value the function can be, or lowest) and test the values you get in d^2/dx^2

In this case you should get a minimum value at x=1, and x=-1. If you plug it back into the original function, you will end up with 2≥2.

If you look at d^2/dx^2 though, you'll see that the function will always be > 0 meaning no matter what extrema you get as a value, it will be concave up. Therefore the function will increase to the left, and to the right of x=1 and x=-1.

since 2 is then your minimum value, and your function x^2+1/x^2 will only increase from there, then it is true that

x^2+1/x^2 will always be greater than or equal to two.

This is equivalent to proving that x^2 + (1/x^2) -2 >= 0.

Multiply through by x^2:

x^4 + 1 - 2x^2 >= 0?

Rearrange:

x^4 - 2x^2 + 1 >= 0?

let w = x^2:

w^2 -2w + 1 >= 0? Where w >0.

The question is: is this equation always greater than or equal to zero for all real values of w > 0? Not too hard to show that it is.

Any number works because -1.999 to 1.999 works and as you go up and down there is no negative possible and x^2 or 1/x^2 will result in a number greater than two.

This is when the 1/x is altogether squared, if not, listen to the answers above.