Prove $\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$ for $0 < x < 1$ I stumbled upon this question  while doing practice inequalities questions, and I do not know how to start...
Problem: Prove that 
\begin{align*}
\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}
\end{align*}
for $0 < x < 1$.
I thought possibly of having an intermediate equality, for example 
\begin{align*}
\sqrt{\frac{2x^2-2x+1}{2}}\ge\text{something}\ge\frac{1}{x+\frac{1}{x}}
\end{align*}
where the "something" is simple, but I could not deduce anything...any help would be appreciated, thanks!

This inequality is true for all $x>0$. Indeed$,$
$$ \frac{(x+\frac{1}{x})^2(2x^2-2x+1)}{2}-1$$
$$={\frac {2 \ \left( 2\ x-1 \right) ^{2}{x}^{2}+ \left( x-1
 \right) ^{4} \left( x+1 \right) ^{2}+{x}^{2} \left( x-1 \right) ^{2}
 \left( x+1 \right) ^{2}}{2{x}^{2}}} \geqq 0$$
