»  Solutions to puzzles in my VDARE.com monthly Diary

  March 2022


In the Math Corner of my March Diary I offered this:

Let a and b be positive integers with a ≥ b. Prove that just one of the following things must be true.

     Either     a/(a+b) + (a+b)/a > √5

     or    a/b + b/a > √5.


•  Solution

If you stare hard there you will see that we are dealing with the function

     f (x) = x + 1/x.

Let's write u for b/a. Then the facts that a and b are positive integers and a ≥ b tell us that u is a positive rational number less than or equal to 1.

With this notation in our hands we can rewrite the thing we're required to prove as:

       For a positive rational number u less than or equal to 1, either  f (u) > √5  or  f (1 + u) > √5.

(Just pause to convince yourself this is right …)

First let's ask:  When is f (x) = √5? That's easy; a simple quadratic equation whose roots turn out to be (√5 ± 1)/2.

Now  (√5 + 1)/2 is of course the Golden Ratio φ, equal to 1.618033988749894848204586834365638117720 …  The other root, (√5 − 1)/2, is φ − 1; and it is also 1/φ (that's what makes the Golden Ratio golden).

Brainteaser At this point it's helpful to have a graph of f (x), so I've drawn one. The red curve is the graph of f (x); the blue horizontal line is y = √5; the purple vertical line is x = φ; the green vertical line is x = φ − 1.

Since we know that u is positive and less than or equal to 1, the only part of the graph that can represent values of u that interest us is the vertical strip from x = 0 (exclusive) to x = 1 (inclusive); and since we know that u is a rational number while φ is irrational, we also know that the green vertical line cannot represent u.

In that strip,  f (u) > √5 when u is between zero (exclusive) and the green line at φ − 1 (exclusive).

But what if u is between the green line at φ − 1 (exclusive) and x = 1 (inclusive)? Well, then 1 + u  is between the purple line at φ (exclusive) and x = 2 (inclusive). For those values, indeed f (1 + u) > √5.  

So for all the values u can take under the conditions given on a and b, either f (u) > √5  or f (1 + u) > √5.

And since u cannot simultaneously be to both the left and the right of the vertical green line, and  f (x) is a single-valued function, that "or" is exclusive: just one of the two things is true. QED.