Prove that

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

I also received this hint: With the square root in the left-hand side, you may be tempted to square both sides, but this gets messy quickly. Instead, you may want to find an intermediate inequality of the form $$\sqrt{\frac{2x^2 - 2x + 1}{2}} \ge \text{something} \ge \frac{1}{x + \frac{1}{x}},$$where "something" is simple.

How should I approach this problem? Should I work backwards? I'm stuck, all solutions are greatly appreciated.


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


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