A Mathematical Rabbit out of an Algebraic Hat

This is how G. D. Chakerian has described his derivation of the 3-term Arithmetic Mean - Geometric Mean Inequality (also Arithmetic-Geometric Mean Inequality or simply AM-GM Inequality) in a chapter of the Mathematical Plums (edited by R. Honsberger.) The inequality asserts that

For any non-negative real numbers $a, b, c,$

(1)

$\sqrt[3]{abc} \le (a + b + c) / 3,$

with equality only if $a = b = c.$

The proof depends on the following identity that holds for non-negative $x, y, z$:

(2)

$\begin{align} x^{3} + y^{3} &+ z^{3} - 3xyz\\ &= \frac{1}{2}(x + y + z)((x - y)^{2} + (y - z)^{2} + (z - x)^{2}) \end{align}$

which is just another form for a more obvious

(3)

$\begin{align} x^{3} + y^{3} + z^{3} &- 3xyz\\ &= (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx) \end{align}$

The right-hand side of the identity in (2) is not negative and so is the left-hand side, implying

$x^{3} + y^{3} + z^{3} \ge 3xyz.$

Now simply substitute $a = x^{3},$ $b = y^{3},$ and $c = z^{3}$ to obtain the AM-GM inequality (1).


Now, for any one who does little mathematics and is not accustomed to problem solving, the derivation, although simple, does appear as a trick out of the blue. Certainly it takes knowledge and experience to come with such a proof. In fact, identity (3) is more basic than (2). The inequality

$x^{2} + y^{2} + z^{2} - xy - yz - zx \ge 0$

can be obtained as the sum of three simpler inequalities

$\begin{align} x^{2} + y^{2} &\ge 2xy,\\ y^{2} + z^{2} &\ge 2yz,\\ z^{2} + x^{2} &\ge 2zx, \end{align}$

which of course explains (2) and might make it less of a mystery.

The identity (2) and similar ones do make frequent appearances in a multitude of settings and are not too difficult to remember.

Problem

(Andreescu et al, p. 9.)

Find all triples $(x, y, z)$ of positive integers such that

$x^{3} + y^{3} + z^{3} - 3xyz = p$

where $p$ is a prime greater than $3.$

The solution comes from (3). Since $p$ is prime and $x + y + z \ge 1$ it must be that $x + y + z = p$ and $x^{2} + y^{2} + z^{2} - xy - yz - zx = 1.$ Now (2) implies that

(4)

$(x - y)^{2} + (y - z)^{2} + (z - x)^{2} = 2.$

Without loss of generality, we may assume $x \ge y \ge z.$ Then from (4), it is either $x = y = z + 1$ or $x - 1 = y = z.$ Prime $p$ has one of the forms $3k + 1$ or $3k + 2.$ In the first case the solutions are $((p+2)/3, (p-1)/3, (p-1)/3);$ in the second case the solutions are $((p+1)/3, (p+1)/3, (p-2)/3).$ For a complete list of solutions, the triples need to be permuted.

References

  1. R. Honsberger, Mathematical Plums, MAA, 1979
  2. T. Andreescu et al, An Introduction to Diophantine Equations, Birkhäuser, 2010

Related material
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  • The Means
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  • Expectation
  • The Size of a Class: Two Viewpoints
  • Averages of divisors of a given integer
  • Family Statistics: an Interactive Gadget
  • Averages in a sequence
  • Arithmetic and Geometric Means
  • Geometric Meaning of the Geometric Mean
  • AM-GM Inequality
  • The Mean Property of the Mean
  • Harmonic Mean in Geometry
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