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 {a_{1}, a_{2},..., a_{n}} =
nth root of a_{1}×+a_{2}×...×a_{n
}In particular the geometric mean of two numbers is
sqrt(a_{1}×a_{2})

An
important result is that the arithmetic mean is always greater than
the geometric mean. For two variables:

(a_{1
}+a_{2
})/2 ≥ sqrt(a_{1}×a_{2})
and attains equality only when a_{1 }=
a_{2 }

This is easily proved.

(a_{1
}- a_{2
})^{2} ≥ 0 and equals zero only when a_{1
}= a_{2 }a_{1
}^{2}+
a_{2 }^{2} – 2 a_{1}a_{2}
≥ 0 and equals zero only when
a_{1 }=
a_{2 }Adding 4a_{1}a_{2}
to both sides gives:

a_{1 }^{2}+
a_{2 }^{2} + 2 a_{1}a_{2}
≥ 4a_{1}a_{2
}

(a_{1 }+ a_{2
})^{2} ≥
4a_{1}a_{2
}

(a_{1 }+
a_{2 }) ≥
2sqrt(a_{1}a_{2}),
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 X^{2} and 1/X^{2} also gives a
minimum as X = 1, so

X + 1/X + X^{2} + 1/X^{2 }is
minimal for X = 1