Use of Lagrange multipliers in pure math problems I realize that Lagrange multipliers are extremely useful for applied optimization problems. 
However, I know that the standard analytic proof of the spectral theorem relies on them. I've also seen a few other uses/mentions of them in some pure math textbooks. (For example, Wade's An Introduction to Analysis uses an exercise on Lagrange multipliers to later prove a result due to Bernstein on the convergence of Fourier series.) 

My question, then, is if Lagrange multipliers are generally a useful
  technique for extremal problems that arise in pure math. If so, are
  there some well-known proofs in this area that use them (other than
  those I mentioned)?

I'm simply curious as to their use outside applied optimization, since the derivation of the existence of the so-called Lagrange multiplier is really just a corollary of a very geometric fact--namely that the gradient is perpendicular to the level sets.
EDIT: To be a bit more specific, by "useful" I mean that it is in fact applicable to certain pure math problems with somewhat regular frequency.
 A: Once, I had to solve this problem:

Given two real numbers $a>b>0$, consider the hyperbola$$H=\{(x,y)\in\mathbb{R}^2\,|\,x^2-y^2=a^2-b^2\}.$$One of its points is $P=(-a,-b)$.
  What is the distance from $P$ to the branch of $H$ to which $P$ does not belong (which, in this case, is the right branch of $H$).

I solved it using Lagrange multipliers. This led me to the system$$\left\{\begin{array}{l}x+a=2\lambda x\\y+b=-2\lambda y\\x^2-y^2=a^2-b^2\\x>0\text{.}\end{array}\right.$$This, in turn, led me to the equation$$\frac{4(a^2-b^2)\lambda^4-3(a^2-b^2)\lambda^2-(a^2+b^2)\lambda}{(1-2\lambda)^2(1+2\lambda)^2}=0\text{.}$$After dividing the numerator by $4(a^2-b^2)\lambda$ (the solution $\lambda=0$ is irrelevant here), one gets a third degree polynomial:$$\lambda^3-\frac34\lambda-\frac{a^2+b^2}{4(a^2-b^2)}.$$Since there is no second degree term, Cardano's formula can be applied directly, giving$$\lambda=\frac12\left(\sqrt[3]{\frac{a-b}{a+b}}+\sqrt[3]{
\frac{a+b}{a-b}}\right)\tag1$$and therefore the point of the right branch of the hyperbola closest to $P$ is $\bigl(-\frac a{1-2\lambda},-\frac b{1+2\lambda}\bigr)$, where the value of $\lambda$ is the one given by $(1)$.
A: I hope the following example is "pure enough". 

Assume $f(x)$ is continuous on $\mathbb{R}$, nonnegative and satisfies $$\int_{-\infty}^{\infty} f(x) dx = 1$$
  If $[a,b]$ is an interval with shortest length such that $$\int_{a}^b f(x) dx = \frac{1}{2}$$ Prove that $f(a)=f(b)$.

Define $F(x,y) = \int_x^y f(t) dt$, then $F$ is of $C^1$ on $\mathbb{R}^2$. We wish to minimize $y-x$ subject to constrain $F(x,y)=1/2$. Then, 
$$(-1, 1) = \nabla (y-x) = \lambda \nabla (F(x,y)) = (-f(x), f(y))$$
Thus, if $[a,b]$ is such that $b-a$ is minimized, we have
$$\lambda f(a) = \lambda f(b) = 1$$
This implies $\lambda \neq 0$, hence $f(a) = f(b) = 1/\lambda$.
A: You can use Lagrange multipliers for the proof of Hadamard's inequality. It says that for a square matrix $A$ with column vectors ${\bf a}_k$ one has
$$\left|\det(A)\right|\leq\prod_{k=1}^n|{\bf a}_k|\ .$$
The claim can be written in the form
$$-1\leq f({\bf a}_1,\ldots,{\bf a}_n)\leq1\ ,\tag{1}$$
whereby I have considered the determinant $f$ as a function of $n$ vector variables, and it is assumed that all $|{\bf a}_k|=1$. In other words, we have the $n$ constraints
$$F_k({\bf a}_1,\ldots,{\bf a}_n):={1\over2}\bigl(a_{1k}^2+a_{2k}^2+\ldots+a_{nk}^2-1\bigr)=0\qquad(1\leq k\leq n)\ ,$$
each of them reigning over just $n$ of the $n^2$ real variables $a_{ik}$. Lagrange's principal function then is
$$\Phi=f-\sum_{k=1}^n\lambda_k F_k\ ,$$
and we have to differentiate $\Phi$ with respect to all variables $a_{ik}$. We obtain the $n^2$ equations
$$\left({\partial \Phi\over\partial a_{ik}}=\right)\quad A_{ik}-\lambda_k a_{ik}=0\qquad({\rm all}\ i,  k)\ ,\tag{2}$$
where $A_{ik}$ is the cofactor of $a_{ik}$ in the determinant $f$.
For the rest of the proof some linear algebra is used. Multiply $(2)$ by $a_{ir}$, and sum over $i$:
$$\sum_{i=1}^n a_{ir}A_{ik}-\lambda_k\>{\bf a}_r\cdot{\bf a}_k=0\qquad ({\rm all}\ r,  k)\ ,$$
which is
$$\lambda_k\>{\bf a}_r\cdot{\bf a}_k=\left\{\eqalign{\det(A)\quad&(r=k)\cr 0\qquad&(r\ne k)\cr}\right.\qquad ({\rm all}\ r,  k)\ .$$
Letting $r=k$ we see tha all $\lambda_k$ have the same value $\det(A)\ne0$ in the extremal situation, and letting $r\ne k$ we then see that ${\bf a}_r\cdot{\bf a}_k=0$ in this case. We therefore have $A'\cdot A=$ identity matrix, and this implies $|\det(A)|=1$.
