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

2 Answers
May 18, 2017

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 #(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#.

May 18, 2017

Answer:

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