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Ishaan94
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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.
 2 years ago
 2 years ago
Ishaan94 Group Title
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.
 2 years ago
 2 years ago

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KingGeorge Group TitleBest 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 \[\frac{1}{y}+ \frac{1}{4z}\geq1\]
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
Thanks. I will try working on it now.
 2 years ago

KingGeorge Group TitleBest 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 \(x<(4z)\).
 2 years ago

Ishaan94 Group TitleBest ResponseYou've already chosen the best response.1
@KingGeorge didn't get the solution :(
 2 years ago

KingGeorge Group TitleBest 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.
 2 years ago

KingGeorge Group TitleBest 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.
 2 years ago

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

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

KingGeorge Group TitleBest 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 \[\frac{5}{2}\leq \frac{y}{y1}\]
 2 years ago

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

KingGeorge Group TitleBest 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 \(z>15/8\).
 2 years ago

KingGeorge Group TitleBest 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 \(z\geq2\).
 2 years ago

Ishaan94 Group TitleBest 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.
 2 years ago

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