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

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KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1Best way to show this, would be to suppose WLOG \[\frac{1}x + \frac1{4y},\;\frac{1}z + \frac1{4x}\]are both smaller than 1, and show that this implies \[\frac{1}{y}+ \frac{1}{4z}\geq1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks. I will try working on it now.

KingGeorge
 4 years ago
Best ResponseYou've already chosen the best response.1Suppose those first two are smaller than 1. That means \(x>1\) and \(y<(4x)\). Also, \(z<1\) and \(x<(4z)\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@KingGeorge didn't get the solution :(

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1I think I messed up on the direction of some of the inequalities. Let me retype that thing real quick.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1We 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.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1and \(z<4\) of course.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Another way to finish the proof would be to show \(z\geq3\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Since you can rearrange the topmost row to get \(z>5/2\), you can also now finish by showing that \[\frac{5}{2}\leq \frac{y}{y1}\]

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1In other words, if \(y\leq 5/3\), we are done.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Hence, 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 \(z>15/8\).

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1In 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 \(z\geq2\).

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I found another way, sum up the three terms and assume their sum to be less than 3. Proof by contradiction.

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1That is also an excellent way one could do this.
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