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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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KingGeorgeBest ResponseYou've already chosen the best response.1
Best 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 \[\frac{1}{y}+ \frac{1}{4z}\geq1\]
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
Thanks. I will try working on it now.
 one year ago

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

Ishaan94Best ResponseYou've already chosen the best response.1
@KingGeorge didn't get the solution :(
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I think I messed up on the direction of some of the inequalities. Let me retype that thing real quick.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
and \(z<4\) of course.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Another way to finish the proof would be to show \(z\geq3\).
 one year ago

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

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

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \(z>15/8\).
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
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 \(z\geq2\).
 one year ago

Ishaan94Best ResponseYou've already chosen the best response.1
I found another way, sum up the three terms and assume their sum to be less than 3. Proof by contradiction.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
That is also an excellent way one could do this.
 one year ago
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