Question

Prove that `\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}`for `0 < x < 1.` ?

Answer 1

See explanation...

Explanation:

Let's take a look at the left hand expression to understand it's behaviour better:

`sqrt((2x^2-2x+1)/2) = sqrt((4x^2-4x+1+1)/4`

`color(white)(sqrt((2x^2-2x+1)/2)) = sqrt(((2x-1)^2+1)/4`

`color(white)(sqrt((2x^2-2x+1)/2)) = 1/2sqrt((2x-1)^2+1)`

Note that since it is a square, we have `(2x-1)^2 >= 0` for all real values of `x`, attaining its minimum value when `x=1/2` and `(2x-1) = 0`.

Hence we find:

`sqrt((2x^2-2x+1)/2) = 1/2sqrt((2x-1)^2+1) >= 1/2sqrt(1) = 1/2`

Now let's look at the right hand expression:

`(x-1)^2 = x^2-2x+1`

`color(white)((x-1)^2) = (x^2+1)-2x`

`color(white)((x-1)^2) = (x^2+1)(1-(2x)/(x^2+1))`

`color(white)((x-1)^2) = 2(x^2+1)(1/2-x/(x^2+1))`

`color(white)((x-1)^2) = 2(x^2+1)(1/2-1/(x+1/x))`

Now `(x-1)^2 >= 0`, attaining its minimum value `0` when `x=1`.

Also `(x^2+1) >= 1`

Hence we find:

`1/2-1/(x+1/x) >= 0`

That is `1/(x+1/x) <= 1/2`, attaining its maximum value `1/2` when `x=1`.

So:

`sqrt((2x^2-2x+1)/2) >= 1/2 >= 1/(x+1/x)`

for all real values of `x`

In fact:

`sqrt((2x^2-2x+1)/2) > 1/(x+1/x)`

since there is no value of `x` for which both sides are equal to `1/2`.

Answer 2

See below.

Explanation:

We know that `x+1/x ge 2` and

`(2x^2-2x+1)/2 = x(x-1)+1/2 le 1/2` for `x in[0,1]`

so

`sqrt(1/2) ge sqrt(x(x-1)+1/2) ge 1/2`

which is true