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

##### 2 Answers

#### Answer:

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

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

Also

Hence we find:

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

That is

So:

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

for all real values of

In fact:

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

since there is no value of

#### Answer:

See below.

#### Explanation:

We know that

so

which is true