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

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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 \[\frac{1}{y}+ \frac{1}{4-z}\geq1\]
Thanks. I will try working on it now.
Suppose those first two are smaller than 1. That means \(x>1\) and \(y<(4-x)\). Also, \(z<1\) and \(x<(4-z)\).

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@KingGeorge didn't get the solution :(
I think I messed up on the direction of some of the inequalities. Let me retype that thing real quick.
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
and \(z<4\) of course.
Another way to finish the proof would be to show \(z\geq3\).
Since you can rearrange the topmost row to get \(z>5/2\), you can also now finish by showing that \[\frac{5}{2}\leq \frac{y}{y-1}\]
In other words, if \(y\leq 5/3\), we are done.
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 \(z>15/8\).
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 \(z\geq2\).
I found another way, sum up the three terms and assume their sum to be less than 3. Proof by contradiction.
That is also an excellent way one could do this.

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