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  "body": "\n## Proof\n\nWe write the directional derivative as a dot product with the gradient and compare it with its largest possible value.\n\n1. Let $U$, $f$ and $a$ be as in the statement. Put $g = \\nabla f(a)$, so $g \\ne 0$. Let $u \\in \\mathbb{R}^n$ with $\\|u\\| = 1$, and put $w = g / \\|g\\|$, a unit vector.\n2. Since [[the derivative gives the directional derivatives|0A6PMR20N4]], the [[directional derivative|0Y7PVW4Q44]] $D_u f(a)$ exists and equals $Df(a)(u)$.\n3. By the definition of the [[gradient|0NS2QYVC4Y]], $Df(a)(u) = g \\cdot u$. So $D_u f(a) = g \\cdot u$.\n4. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]] in [[Euclidean space|0XPR1E85W0]], $|g \\cdot u| \\le \\|g\\| \\, \\|u\\| = \\|g\\|$. So $-\\|g\\| \\le D_u f(a) \\le \\|g\\|$.\n5. Expanding the dot product, $\\|u - w\\|^2 = \\|u\\|^2 - 2 \\, u \\cdot w + \\|w\\|^2 = 2 - 2 \\, \\dfrac{g \\cdot u}{\\|g\\|}$.\n6. In the same way, $\\|u + w\\|^2 = 2 + 2 \\, \\dfrac{g \\cdot u}{\\|g\\|}$.\n7. By steps 3 and 5, $D_u f(a) = \\|g\\|$ holds if and only if $\\|u - w\\| = 0$, that is, if and only if $u = g / \\|g\\|$.\n8. By steps 3 and 6, $D_u f(a) = -\\|g\\|$ holds if and only if $\\|u + w\\| = 0$, that is, if and only if $u = -g / \\|g\\|$.\n9. So the upper bound is attained exactly at $u = \\nabla f(a) / \\|\\nabla f(a)\\|$, and the lower bound exactly at $u = -\\nabla f(a) / \\|\\nabla f(a)\\|$.\n\n## Idea\n\nThe slope of $f$ in a unit direction $u$ is the dot product $\\nabla f(a) \\cdot u$. A dot product with a fixed vector is largest when $u$ points the same way as that vector, and smallest when it points the opposite way. The identity $\\|u - w\\|^2 = 2 - 2 u \\cdot w$ for unit vectors makes the cases of equality exact.\n",
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