# A night with Cut the Knot, Nassim Taleb, and some Supernova

Please note the correction at the bottom of the post

A further correction – there is still an error. Ugh. This approach may not work, unfortunately . . .

Saw a neat problem from Alexander Bogomolny earlier today:

I actually missed the problem when it was initial posted, but saw it via Nassim Taleb’s solution:

The problem sort of gnawed at me all day and I figured it was in the maybe 1 in 10 problems that Bogomolny posts that I might be able to solve.

So, tonight I poured a glass of Supernova and gave it a go One thing on my mind all day with this problem was Jensen’s inequality. What I would *love* to be able to do is say that by Jensen’s inequality: $(1/3) \sqrt{x^2 + 3} + (1/3) \sqrt{y^2 + 3} + (1/3) \sqrt{xy + 3}$ $\geq \sqrt{ (1/3)( x^2 + y^2 + xy) + 3}$

Which is easily seen to be $\geq 2$ because of the constraint $x + y = 2.$ That work would show the original inequality was $\geq$ 6.

The approach has a tiny bit of merit since $\sqrt{x^2 + 3}$ is concave up for $x$ between 0 and 2 -> here’s a little Mathematica plot showing that the second derivative is indeed positive on 0 to 2: But . . . the problem is that folding in the 3rd term in the sum is stretching the rules of Jensen’s inequality a bit, I think, since it is not of the form $\sqrt{a^2 + 3}$.

With the first two terms, though, applying Jensen’s inequality seems ok, but I now need (1/2)’s instead of (1/3)’s since there are only two terms. So, I’ll use Jensen’s on the first two terms only and try to show that $(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3$

By Jensen’s inequality this new sum is $\geq \sqrt{ \frac{x^2 + y^2}{2} + 3} + (1/2) \sqrt{xy + 3}$

A bit of algebra and the fact that $x + y = 2$ allows us to simplify this expression to: $\sqrt{5 - xy} + (1/2) \sqrt{xy + 3}$

and then further to: $\sqrt{ (x - 1)^2 + 4} + (1/2) \sqrt{4 - (x - 1)^2}$

Now we are just down to a fairly straightforward calculus problem, and I’ll let Mathematica do the heavy lifting since the algebra isn’t that interesting: We can see visually that the minimum occurs at $x = 1$ from the plot, and the plot of the derivative further confirms that there is only one critical point. The value of the last expression at $x = 1$ is indeed 3 as we were hoping.

So, Jensen’s inequality, a bit of calculus, and a nice glass of scotch shows that the original inequality is indeed true.

Thanks to Alexander Bogomolny for the problem, and to Nassim Taleb for his solution that got me thinking about the problem.

Correction

I received a note from Alexander Bogomoly over night. He spotted an error in the calculation:

and I thought my kids having trouble sleeping and waking me up at 5:00 am today was a bad start to the day!

But it seems that I’ve gotten very lucky – both learning from my carelessness in applying Jensens inequality and that the path forward from Bogomolny’s correction is easier than the path I actually took.

Starting here – we wish to show that: $(1/2) \sqrt{x^2 + 3} + (1/2) \sqrt{y^2 + 3} + (1/2) \sqrt{xy + 3} \geq 3$

The correction shows that the expression on the left hand side is $\geq$ than $\sqrt{ (\frac{x + y}{2})^2 + 3} + (1/2) \sqrt{xy + 3}$

but since $x + y = 2$, the first piece of this expression is equal to 2 and the 2nd expression simplifies as before. So we are left with $2 + (1/2) \sqrt{4 - (x - 1)^2}$

and this expression has a maximum of 3 at $x = 1.$

That means that the expression we were trying to show to be greater than 3 is indeed greater than 3, and the expression in the original tweet is greater than 6.

I’m grateful to Alexander Bogomolny for spotting the error in my original argument.