## zmudz one year ago Prove that $$\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}$$ for $$0 < x < 1.$$

1. thomas5267

Square both side first?

2. anonymous

Fix up the RHS first.

3. anonymous

Note that $$x+1/x \geq 2$$ (by arithmetic mean > geometric mean or simply by completing the square) Also, $$2x^2 - 2x + 1 = (1-x)^2 + x^2$$ Now, use the fact that the root-means square is greater than the arithmetic mean, $$\sqrt{\frac{2x^2 - 2x + 1}{2}} = \sqrt{\frac{(1-x)^2 + x^2}{2}} \geq \frac{(1-x) + x}{2} = \frac{1}{2}$$ Multiply the above two inequalities to get what is required.

4. Loser66

@adxpoi I would like to know about the so-called "arithmetic mean> geometric mean". Would you please explain me?

5. Loser66

The problem is easily proving by "square both sides" and do backward of the flow. I would like to learn a new method.

6. anonymous

Arithmetic mean is$\frac{ x_1+x_2+x_3+...+x_n }{ n }$while geometric mean is$\sqrt[n]{x_1x_2x_3...x_n}$

7. thomas5267
8. anonymous

Sure. The AM > GM inequality states that for $$n$$ non-negative real numbers $$x_1, x_2,..., x_n$$ the following holds: $\frac{x_1+x_2+...+x_n}{n} \geq (x_1.x_2....x_n)^{\frac{1}{n}}$

9. Loser66

Thank you. I will read it carefully. @thomas5267

10. anonymous

A possible proof is by induction, first prove it for two numbers, easily done. Then try to extend it by induction.

11. Loser66

@adxpoi Yes, I will. Thank you.

12. ganeshie8

Alternatively you may show that the max value of right hand side is $$\frac{1}{2}$$ and min value of left hand side is $$\frac{1}{2}$$. They occur at different places, so the inequality follows.

13. ganeshie8

thats exactly same as your first proof, but I was looking at it more from vertex of quadratic equation side.. :)

14. Loser66

@ganeshie8 I looooooove your max/min method.!! Wow!!!

15. anonymous

Cumbersome, but doesn't require knowledge of the theorems mentioned above.$\sqrt{\frac{ 2x^2-2x+1 }{ 2 }} \ge \frac{ 1 }{ x+\frac{ 1 }{ x } }$$\sqrt{\frac{ 2\left( x^2-x+\frac{ 1 }{ 2 } \right) }{ 2 }} \ge \frac{ x }{ x^2+1 }$$x^2-x+\frac{ 1 }{ 2 } \ge \frac{ x^2 }{ x^4+2x^2+1 }$$x^6-x^5+\frac{ 5 }{ 2 }x^4-2x^3+x^2-x+\frac{ 1 }{ 2 } \ge 0$$\left( x^6+\frac{ 5 }{ 2 }x^4+x^2+\frac{ 1 }{ 2 } \right) - \left( x^5+2x^3+x \right) \ge 0$$x=0 \rightarrow \frac{ 1 }{ 2 } \ge 0$$x=1 \rightarrow 0 \ge 0$

16. anonymous

Excuse me. Last line above $x=1 \rightarrow 1 \ge 0$should read

17. anonymous

@ganeshie8 isn't the minimum of the LHS $$\sqrt\frac{1}{2}$$ ? If so, how does that affect the max/min proposition?

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