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!

  • $\begingroup$ Note that $2x^2-2x+1=x^2+(x^2-2x+1)=x^2+(x-1)^2$, maybe you can use A.G. inequality. $\endgroup$
    – DiegoMath
    Oct 24, 2014 at 3:14

2 Answers 2



Notice that $\sqrt{2x^2-2x+1}=\sqrt{x^2+(x-1)^2} \ge \sqrt{(\frac{x+1-x}{2})^2}=\frac12$ and $\frac{1}{x+\frac1x}\le \frac{1}{2x\frac1x}=\frac12$


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

$$={\frac {2 \ \left( 2\ x-1 \right) ^{2}{x}^{2}+ \left( x-1 \right) ^{4} \left( x+1 \right) ^{2}+{x}^{2} \left( x-1 \right) ^{2} \left( x+1 \right) ^{2}}{2{x}^{2}}} \geqq 0$$


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