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Prove that

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

I've tried to find some intermediate expression where $\sqrt{\frac{2x^2 - 2x + 1}{2}} \ge \text{something} \ge \frac{1}{x + \frac{1}{x}}$, but I can't figure out how to get it working. Squaring both sides will only create a huge mess.

Any help would be greatly appreciated. Thanks!

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We have: $(a+b)^2 \le 2(a^2+b^2)\implies \sqrt{\dfrac{a^2+b^2}{2}} \ge \dfrac{a+b}{2}$. Apply this inequality for $a = x, b = 1-x$, and note that $x+\dfrac{1}{x} \ge 2$, and the answer follows.

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  • $\begingroup$ 0<x<1 so x+1/x > 2 $\endgroup$ – user379195 May 12 '17 at 1:44
  • $\begingroup$ Its still true though.... $\endgroup$ – DeepSea May 12 '17 at 1:44
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Hint (it's enough that $x \gt 0\,$, actually):

$$\left(x + \frac{1}{x}\right) \,\sqrt{ \left(x-\frac{1}{2}\right)^2+\frac{1}{4} } \;\geq\; 2 \cdot \sqrt{\frac{1}{4}} \,=\, 1$$

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