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.

Quote:

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Originally Posted by Ross321

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.

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This would be my approach.
First, find and

set (this will get you either the highest value the function can be, or lowest) and test the values you get in

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 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 will only increase from there, then it is true that

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.