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

May 18, 2017

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

$\textcolor{w h i t e}{\sqrt{\frac{2 {x}^{2} - 2 x + 1}{2}}} = \frac{1}{2} \sqrt{{\left(2 x - 1\right)}^{2} + 1}$

Note that since it is a square, we have ${\left(2 x - 1\right)}^{2} \ge 0$ for all real values of $x$, attaining its minimum value when $x = \frac{1}{2}$ and $\left(2 x - 1\right) = 0$.

Hence we find:

$\sqrt{\frac{2 {x}^{2} - 2 x + 1}{2}} = \frac{1}{2} \sqrt{{\left(2 x - 1\right)}^{2} + 1} \ge \frac{1}{2} \sqrt{1} = \frac{1}{2}$

Now let's look at the right hand expression:

${\left(x - 1\right)}^{2} = {x}^{2} - 2 x + 1$

$\textcolor{w h i t e}{{\left(x - 1\right)}^{2}} = \left({x}^{2} + 1\right) - 2 x$

$\textcolor{w h i t e}{{\left(x - 1\right)}^{2}} = \left({x}^{2} + 1\right) \left(1 - \frac{2 x}{{x}^{2} + 1}\right)$

$\textcolor{w h i t e}{{\left(x - 1\right)}^{2}} = 2 \left({x}^{2} + 1\right) \left(\frac{1}{2} - \frac{x}{{x}^{2} + 1}\right)$

$\textcolor{w h i t e}{{\left(x - 1\right)}^{2}} = 2 \left({x}^{2} + 1\right) \left(\frac{1}{2} - \frac{1}{x + \frac{1}{x}}\right)$

Now ${\left(x - 1\right)}^{2} \ge 0$, attaining its minimum value $0$ when $x = 1$.

Also $\left({x}^{2} + 1\right) \ge 1$

Hence we find:

$\frac{1}{2} - \frac{1}{x + \frac{1}{x}} \ge 0$

That is $\frac{1}{x + \frac{1}{x}} \le \frac{1}{2}$, attaining its maximum value $\frac{1}{2}$ when $x = 1$.

So:

$\sqrt{\frac{2 {x}^{2} - 2 x + 1}{2}} \ge \frac{1}{2} \ge \frac{1}{x + \frac{1}{x}}$

for all real values of $x$

In fact:

$\sqrt{\frac{2 {x}^{2} - 2 x + 1}{2}} > \frac{1}{x + \frac{1}{x}}$

since there is no value of $x$ for which both sides are equal to $\frac{1}{2}$.

May 18, 2017

See below.

Explanation:

We know that $x + \frac{1}{x} \ge 2$ and

$\frac{2 {x}^{2} - 2 x + 1}{2} = x \left(x - 1\right) + \frac{1}{2} \le \frac{1}{2}$ for $x \in \left[0 , 1\right]$

so

$\sqrt{\frac{1}{2}} \ge \sqrt{x \left(x - 1\right) + \frac{1}{2}} \ge \frac{1}{2}$

which is true