2025-09-03

So we have some inequality in the variables \((x_1, x_2, x_3,\dots, x_n)\) and we are asked to find the minimum or the maximum. Then we look at the equality case - suppose the extreme occurs at \((y_1, y_2, y_3, \dots, y_n)\). We will try to find an operation \(P\) such that

  1. The inequality \(f(x_1, x_2, \dots, x_n) \ge f(P(x_1, x_2, \dots, x_n))\) holds.
  2. \(P(x_1, x_2, \dots, x_n)\) preserves conditions regarding \((x_1, x_2, \dots, x_n)\). For example, \(x_1 + x_2 +\dots + x_n = 1\) or \(x_1x_2 \dots x_n = 1\) should be preserved.
  3. The repeated operation of \(P\) on any initial tuple \((x_1, x_2, \dots, x_n)\) results in \((y_1, y_2, \dots, y_n)\) or similar (since for upto two free variables it is not hard to solve an inequality).

1999 China TST

For non-negative reals \(x_1\), \(x_2\), \(\dots\), \(x_n\) such that \(x_1 + x_2 + \dots + x_n = 1\), find the maximum of \[\sum_{j=1}^{n} (x_j^4 - x_j^5)\]

Metagaming, maybe the maximum is at \((1/n, 1/n, \dots 1/n)\) and equal to \((n-1)/n^4\). But this is very far from true! Consider the case \((1/2, 1/2, 0, 0, \dots, 0)\), where \(1/16\) is achieved, far bigger than \((n-1)/n^4\).

But there is a problem. The maximum value for \(n=2\) does not occur for \((1/2, 1/2)\) but for \((\frac{3 - \sqrt{3}}{6}, \frac{3 + \sqrt{3}}{6})\), and the value is \(1/12\). Surely the only possibility now is that the maximum for \(n\) happens at \((\frac{3 - \sqrt{3}}{6}, \frac{3 + \sqrt{3}}{6}, 0, \dots, 0)\), right?

The only way we get \(0\) through smoothing is by replacing \((x, y) \to (x +y, 0)\). So take some \(x\), \(y\) (nonzero) - we want \[(x + y)^4 - (x + y)^5 \ge x^4 + y^4 - x^5 - y^5 \] \[(x + y)^4 - x^4 - y^4 \ge (x + y)^5 - x^5 - y^5 \implies 2(2x^2+3xy+2y^2) \ge 5(x+y)(x^2+xy+y^2)\] \[x + y \le \frac{2(2x^2+3xy+2y^2)}{5(x^2+xy+y^2)} = \frac{4}{5} + \frac{2xy}{5x^2+5xy+5y^2} \] We cannot increase the RHS after \(4/5\) since the additional term can be arbitrarily close to \(0\) I think. So we get that if \(x + y \le 4/5\), then we can do this merging. But now notice that for any three \(x\), \(y\), \(z\) such that \(x + y + z \le 1\) and all three pairs satisfy \(x + y \ge 4/5\), \(y + z \ge 4/5\), \(z + x \ge 4/5\), then their sum becomes \(\ge 12/5\), which is a contradiction since \(2(x+y+z) \le 2\).

This means that we can perform the operation up till when there are \(2\) nonzero terms. At that point we already know that the maximum is obtained by \((\frac{3 - \sqrt{3}}{6}, \frac{3 + \sqrt{3}}{6})\), so we are done - the maximum is \(1/12\) for all \(n \ge 2\).