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  "body": "\n## Proof\n\nWe apply the theorem with several constraints to the single constraint $g - c$.\n\n1. Let $U$, $n \\ge 2$, $f$, $g$, $c$ and $a$ be as in the statement. Define $G = g - c$ on $U$. It is of [[class C^1|0TCBZMEYGN]], since $g$ is.\n2. The [[level set|0Y5Q8Q9WJ7]] $\\set{x \\in U : g(x) = c}$ is the set $M = \\set{x \\in U : G(x) = 0}$.\n3. The [[gradient|0NS2QYVC4Y]] of $G$ is $\\nabla G = \\nabla g$, since $c$ is constant. So $\\nabla G(a) \\ne 0$.\n4. A single nonzero vector is [[linearly independent|0EVHJFTG9Z]]. Also $m = 1 \u003c n$, since $n \\ge 2$.\n5. The restriction of $f$ to $M$ has a local extremum at $a$. By the [[Lagrange multiplier theorem|0VKKP8X6Q6]] with $m = 1$ and the constraint $G$, there is a unique $\\lambda \\in \\R$ with $\\nabla f(a) = \\lambda \\nabla G(a)$.\n6. By step 3, this says $\\nabla f(a) = \\lambda \\nabla g(a)$, and $\\lambda$ is unique.\n\n## Idea\n\nA level set $g = c$ is the zero set of $g - c$, which has the same gradient as $g$. One nonzero gradient is a linearly independent family, so the general theorem applies with one multiplier.\n",
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