Question

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.

Answer 1

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 \[\frac{1}{y}+ \frac{1}{4-z}\geq1\]

Answer 2

Thanks. I will try working on it now.

Answer 3

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

Answer 4

@KingGeorge didn't get the solution :(

Answer 5

I think I messed up on the direction of some of the inequalities. Let me retype that thing real quick.

Answer 6

Okay.

Answer 7

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.

Answer 8

and \(z<4\) of course.

Answer 9

Another way to finish the proof would be to show \(z\geq3\).

Answer 10

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

Answer 11

In other words, if \(y\leq 5/3\), we are done.

Answer 12

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 \(z>15/8\).

Answer 13

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 \(z\geq2\).

Answer 14

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

Answer 15

That is also an excellent way one could do this.