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zmudz

  • one year ago

Prove that \(\sqrt{ \frac{2x^2 - 2x + 1}{2} } \geq \frac{1}{x + \frac{1}{x}}\) for \(0 < x < 1.\)

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  1. thomas5267
    • one year ago
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    Square both side first?

  2. anonymous
    • one year ago
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    Fix up the RHS first.

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

  5. Loser66
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    https://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means

  8. anonymous
    • one year ago
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    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
    • one year ago
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    Thank you. I will read it carefully. @thomas5267

  10. anonymous
    • one year ago
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    A possible proof is by induction, first prove it for two numbers, easily done. Then try to extend it by induction.

  11. Loser66
    • one year ago
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    @adxpoi Yes, I will. Thank you.

  12. ganeshie8
    • one year ago
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    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
    • one year ago
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    thats exactly same as your first proof, but I was looking at it more from vertex of quadratic equation side.. :)

  14. Loser66
    • one year ago
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    @ganeshie8 I looooooove your max/min method.!! Wow!!!

  15. anonymous
    • one year ago
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    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
    • one year ago
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    Excuse me. Last line above \[x=1 \rightarrow 1 \ge 0\]should read

  17. anonymous
    • one year ago
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    @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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