Prove that #\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}#for #0 < x < 1.# ?
2 Answers
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
See below.
Explanation:
We know that
so
which is true