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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!

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  • $\begingroup$ 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 '14 at 3:14
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Hints:

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

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Set \begin{aligned} P = \left(x + \frac{1}{x}\right)^2\cdot\frac{2x^2 - 2x + 1}{2} -1. \end{aligned} We have \begin{aligned} P= \frac{2(x^4+2x^2)(x-1)^2+x^2(x^2-x+1)^2+(x^2+2)(x^2+x-1)^2}{4x^2} \geqslant 0. \end{aligned}

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