For high school math olympiad students, OTIS XII applications are out, due August 1, 2026.

But I’m really going to use that short announcement as an excuse to describe the second most common error I see on the OTIS application (the most common error is this type-error I wrote about in 2023). The mistake is so obvious I feel stupid writing about it, but I keep seeing it. So I’m going to record it anywayFree blog advertisement every time I see this mistake again, right?, even though most of the readers of the blog will laugh at me.

Here’s a problem I made up to demonstrate it:

Problem: For real numbers $a$ and $b$, determine the minimum possible value of $$f(a,b) = (ab-1)^2 + (a-3)^2 + (b-3)^2.$$

And here’s the wrong solution I keep seeing from students.In recent years, I’ve also seen LLM’s make the same mistake with rage-inducing frequency. But this mistake was common before LLM’s became popular.

Nonsense solution: Because the inequality is symmetric, the minimum must occur when $a=b$.

By (insert some method) the minimum value of $(x^2-1)^2 + 2(x-3)^2$ occurs at $x = \sqrt[3]{3}$. This gives an answer of $$f\left(\sqrt[3]{3}, \sqrt[3]{3}\right) = 19 - 9\sqrt[3]{3} \approx 6.0198.$$

There’s a variant of this where the student tries to use calculus or Lagrange multipliers or whatever first, and thus gets some equations describing the critical points.In this variant, there’s also the issue of critical points ≠ global extrema, see my Lagrange multiplier notes. Then they assert that because the equations are symmetric, the solutions to the equations (and hence, the critical points) are symmetric too.

Of course, neither of these symmetry citations is true.And, I feel like there’s really an obvious sanity check not passing here. If every symmetric inequality really achieved optimums when $a=b=c$, wouldn’t we just cite this on every such problem?

This was the simplest example I could come up with that provides a blatant counterexample to the hopeful symmetry claim. (It was actually surprisingly tricky for me to design a counterexample that wasn’t too complicated and had nice numbers in it.) The true minimum occurs at $f(1,2) = f(2,1) = 6$. And yes, there’s a short non-calculus proof; see if you can find it.