Another Approach to Minimum Problem

Nick Hobson sent this way of solving the problem, which requires neither calculus nor the assumption that the minimum is attained only once. Nick has his own math Web site at Nick's Site, which includes an extensive and informative collection of problems.

Nick's solution requires use of a well known result about arithmetic and geometric means.

The arithmetic mean of a set of numbers is simply what we normally consider their average. The geometric mean is another type of average which is sometimes used. The geometric mean of n numbers is defined as the nth root of their product.

The geometric mean of {a₁, a₂,..., a_n} = nth root of a₁×+a₂×...×a_n
In particular the geometric mean of two numbers is sqrt(a₁×a₂)

An important result is that the arithmetic mean is always greater than the geometric mean. For two variables:
(a₁ +a₂ )/2 ≥ sqrt(a₁×a₂) and attains equality only when a₁ = a₂

This is easily proved.
(a₁ - a₂ )² ≥ 0 and equals zero only when a₁ = a₂
a₁ ²+ a₂ ² – 2 a₁a₂ ≥ 0 and equals zero only when a₁ = a₂
Adding 4a₁a₂ to both sides gives:
a₁ ²+ a₂ ² + 2 a₁a₂ ≥ 4a₁a₂
(a₁ + a₂ )² ≥ 4a₁a₂
(a₁ + a₂ ) ≥ 2sqrt(a₁a₂), and the result follows

Applying this result to X and 1/X, we have
(X + 1/X)/2 ≥ (X * 1/X)
(X + 1/X) ≥ 2 and is equal to 2 only when X = 1/X.
That is, the function attains its minimum value of 2 at X = 1.

A similar analysis for X² and 1/X² also gives a minimum as X = 1, so
X + 1/X + X² + 1/X² is minimal for X = 1