Solving IMO 1984's Easiest Math Problem

Prove that

0 \leq yz + zx + xy - 2xyz \leq \dfrac{7}{27}

, where x, y, z are positive real numbers for which

x + y + z = 1

. This problem comes from the line-up at IMO 1984.

The problem we’re looking at today was the easiest of the 6 problems given at the International Math Olympiad, in 1984.

To solve this problem, we will employ the use of the HM-GM-AM-QM inequality.

As a refresher, for any set of n positive numbers we have: $HM \leq GM \leq AM \leq QM$ , where:

HM (harmonic mean): is n divided by the sum of the reciprocals of the numbers.
GM (geometric mean): is the nth root of the product of the numbers
AM (arithmetic mean): is the sum of the numbers divided by n
QM (quadratic mean): is the square root of the arithmetic mean of the squares of the numbers.

We will divide our problem into two complementary cases.

Without loss of generality due to symmetry we can assume $x=0$ .

Replacing x with 0 in our initial inequality we arrive at this new problem:

$0 \leq yz \leq \dfrac{7}{27}, \text{ where } y+z \geq 0, y+z=1$

The left-hand side (LHS) is obvious, as both numbers are positive

For the right-hand side (RHS), we use the AM-GM inequality, which quickly solves it

Knowing that the numbers are strictly positive, we move $2xyz$ to the left side and divide by the product of the numbers.

Simplifying and re-ordering, we arrive at:

$2 \leq \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}$

This is a trivial consequence of AM-HM inequality.

We re-write our inequality as such:

$(1-2x)(1-2y)(1-2z) \leq \dfrac{1}{27}$

We further split this case into two:

In this case, (1-2x), (1-2y), and (1-2z) are positive, so we can apply GM-AMto them; which directly proves our inequality

We know that $(1-2x)(1-2y)(1-2z)<0$ so it is less than $\dfrac{1}{27}$