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Ishaan94

  • 3 years ago

Let \(x,y\) and \(z\) be positive real numbers less than 4. Prove that among the numbers\[\frac{1}x + \frac1{4-y},\;\frac{1}z + \frac1{4-x},\;\frac{1}y + \frac1{4-z}\]there is at least one that is greater than or equal to 1.

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  1. KingGeorge
    • 3 years ago
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    Best way to show this, would be to suppose WLOG \[\frac{1}x + \frac1{4-y},\;\frac{1}z + \frac1{4-x}\]are both smaller than 1, and show that this implies \[\frac{1}{y}+ \frac{1}{4-z}\geq1\]

  2. Ishaan94
    • 3 years ago
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    Thanks. I will try working on it now.

  3. KingGeorge
    • 3 years ago
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    Suppose those first two are smaller than 1. That means \(x>1\) and \(y<(4-x)\). Also, \(z<1\) and \(x<(4-z)\).

  4. Ishaan94
    • 3 years ago
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    @KingGeorge didn't get the solution :(

  5. KingGeorge
    • 3 years ago
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    I think I messed up on the direction of some of the inequalities. Let me retype that thing real quick.

  6. Ishaan94
    • 3 years ago
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    Okay.

  7. KingGeorge
    • 3 years ago
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    We need \(x>1\) and \(3>y\) and \(4−y>\frac{x}{x−1}\). We also need \(z>1\), \(3>x\), and \(4−x>\frac{z}{z−1}\). Combine all of these things together, we get the following criteria.\[1<x<3,\qquad x<4−\frac{z}{z−1}\]\[y<4−\frac{x}{x−1},\qquad y<3\]\[1<z\]And various rearrangements. We want to show one of the following. \[y≤1\]\[ 4−z\leq \frac{y}{y−1}\] There we go. I hope that's finally right.

  8. KingGeorge
    • 3 years ago
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    and \(z<4\) of course.

  9. KingGeorge
    • 3 years ago
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    Another way to finish the proof would be to show \(z\geq3\).

  10. KingGeorge
    • 3 years ago
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    Since you can rearrange the topmost row to get \(z>5/2\), you can also now finish by showing that \[\frac{5}{2}\leq \frac{y}{y-1}\]

  11. KingGeorge
    • 3 years ago
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    In other words, if \(y\leq 5/3\), we are done.

  12. KingGeorge
    • 3 years ago
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    Hence, we must choose a \(y>5/3\). Using this, we find that \(x>7/4\), and \(z>9/5\). Therefore, \(y>11/6\), \(x>13/7\), and \(z>15/8\).

  13. KingGeorge
    • 3 years ago
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    In general, if you continue to do this process off to infinity (fairly straightforward to show), eventually you will find that \(x\geq2\), \(y\geq2\), and \(z\geq2\).

  14. Ishaan94
    • 3 years ago
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    I found another way, sum up the three terms and assume their sum to be less than 3. Proof by contradiction.

  15. KingGeorge
    • 3 years ago
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    That is also an excellent way one could do this.

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