As an interesting example of the Lagrange multiplier method, we employ it to prove the arithmetic-geometric means inequality:

To begin with, define $f\colon \real_{\geq 0}^n \to \real_{\geq 0}$ by $f(x) = (x_1 \dotsm x_n)^{1/n}$ .

Fix $a > 0$ (the arithmetic mean), and consider the set $M = \{ x \in \real^n : x_i \geq 0 { for all $i$, } \sum_i x_i = na \}$ . This is clearly a closed and bounded set in $\real^n$ , and hence is compact. Therefore $f$ on $M$ has both a maximum and minimum.

$M$ is almost a manifold (a differentiable surface), except that it has a boundary consisting of points with $x_i = 0$ for some $i$ . Clearly $f$ attains the minimum zero on this boundary, and on the ``interior'' of $M$ , that is $M' = \{ x \in \real^n : x_i > 0 { for all $i$, } \sum_i x_i = na \}$ , $f$ is positive. Now $M'$ is a manifold, and the maximum of $f$ on $M$ is evidently a local maximum on $M'$ . Hence the Lagrange multiplier method may be applied.

We have the constraint ¹ $0 = g(x) = \sum_i x_i - na$ under which we are to maximize $f$ .

We compute:

Since we have obtained a unique solution to the Lagrange multiplier system, this unique solution must be the unique maximum point of $f$ on $M$ . So $f(x) \leq a$ amongst all numbers $x_i$ whose arithmetic mean is $a$ , with equality occurring if and only if all the $x_i$ are equal to $a$ .

The case $a = 0$ , which was not included in the above analysis, is trivial.

Proof of concavity

Actually, it turns out to be not much harder to show that $f$ is weakly concave everywhere on $\real_{\geq 0}^n$ , not just on $M$ . A plausibility argument for this fact is that the graph of $t \mapsto \sqrt[n]{t}$ is concave, and since $f$ also involves an $n$ th root, it should be concave too. We will prove concavity of $f$ by showing that the Hessian of $f$ is negative semi-definite.

Since $M$ is a convex² set, the restriction of $f$ to $M$ will also be weakly concave; then we can conclude that the critical point $x_i = a$ on $M$ must be a global minimum on $M$ .

We begin by calculating second-order derivatives:

Let be a test vector. Negative semi-definiteness means $v \, \D^2 f(x) \, v^{\mathrm{T}} \leq 0$ for all $v$ . So let us write out:

Moreover, the case of equality in the Cauchy-Schwarz inequality () only occurs when is parallel to , i.e. $v_i = \mu x_i$ for some scalar $\mu$ . Notice that is the normal vector to the tangent space of $M$ , so unless $\mu = 0$ . This means, equality never holds in () if $v \neq 0$ is restricted to the tangent space of $M$ . In turn, this means that $f$ is strictly concave on the convex set $M$ , and so $x_i = a$ is a strict global minimum of $f$ on $M$ .

Proof of inequality by symmetry argument

Observe that the maximum point $x_i = a$ is pleasantly symmetric. We can show that the solution has to be symmetric, by exploiting the symmetry of the function $f$ and the set $M$ .

We have already calculated that $f$ is strictly concave on $M$ ; this was independent of the Lagrange multiplier method. Since $f$ is strictly concave on a closed convex set, it has a unique maximum there; call it $x$ .

Pick any indices $i$ and $j$ , ranging from $1$ to $n$ , and form the linear transformation $T$ , which switches the $i$ th and $j$ th coordinates of its argument.

By definition, $f(x) \geq f(\xi)$ for all $\xi \in M$ . But $f \circ T = f$ , so this means $f(Tx) \geq f(T\xi)$ for all $\xi \in M$ . But $M$ is mapped to itself by the transformation $T$ , so this is the same as saying that $f(Tx) \geq f(\xi)$ for all $\xi \in M$ . But the global maximum is unique, so $Tx = x$ , that is, $x_i = x_j$ .

Note that the argument in this last proof is extremely general -- it takes the form: if $x$ is the unique maximum of $f \colon M \to \real$ , and $T$ is any symmetry of $M$ such that $f \circ T = f$ , then $Tx = x$ . This kind of symmetry argument was used to great effect by Jakob Steiner in his famous attempted proof of the isoperimetric inequality: among all closed curves of a given arc length, the circle maximizes the area enclosed. However, Steiner did not realize that the assumption that there is a unique maximizing curve has to be proved. Actually, that is the hardest part of the proof -- if the assumption is known, then one may easily calculate using the Lagrange multiplier method that the maximizing curve must be a circle!

Footnotes

... constraint¹

Although $g$ does not constrain that $x_i > 0$ , this does not really matter -- for those who are worried about such things, a rigorous way to escape this annoyance is to simply define $f(x) = 0$ to be zero whenever $x_i \leq 0$ for some $i$ . Clearly the new $f$ cannot have a maximum at such points, so we may simply ignore the points with $x_i \leq 0$ when solving the system. And for the same reason, the fact that $f$ fails to be differentiable at the boundary points is immaterial.

... HREF="http://planetmath.org/encyclopedia/Polyconvex.html">convex²

If $M$ is not convex, then the arguments that follow will not work. In general, a prescription to determine whether a critical point is a local minimum or maximum can be found in tests for local extrema in Lagrange multiplier method.