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  "body": "\n## Proof\n\nLet $\\mathbb{K}$ be the scalar field, $\\R$ or $\\C$. Over $\\R$, the [[complex conjugate|0BJV24C4SY]] $\\bar a$ of a scalar $a$ is $a$ itself.\n\n1. By axiom 1 of the [[inner product|0NB0AWGJBW]], $\\langle 0, v \\rangle = \\langle 0 \\cdot 0, v \\rangle = 0 \\langle 0, v \\rangle = 0$ for every $v \\in V$.\n2. By axioms 1 and 2, for $u, v, w \\in V$ and $a, b \\in \\mathbb{K}$, we get $\\langle w, a u + b v \\rangle = \\overline{a \\langle u, w \\rangle + b \\langle v, w \\rangle}$. This equals $\\bar a \\langle w, u \\rangle + \\bar b \\langle w, v \\rangle$.\n3. By axiom 2, $\\langle v, v \\rangle$ equals its own conjugate, so it is real. By axiom 3 and step 1, $\\langle v, v \\rangle \\ge 0$, with equality only for $v = 0$.\n4. So the [[induced norm|0MFJJV97RJ]] $\\norm{v} = \\sqrt{\\langle v, v \\rangle}$ is a real number $\\ge 0$. Also $\\norm{v} = 0$ holds if and only if $v = 0$.\n5. Suppose first that $y = 0$. By steps 1 and 2, $\\langle x, 0 \\rangle = \\overline{\\langle 0, x \\rangle} = 0$, and $\\norm{y} = 0$. So both sides are $0$ and equality holds.\n6. In this case $x$ and $y$ are also linearly dependent, since $0 \\cdot x + 1 \\cdot y = 0$. So the whole claim holds when $y = 0$.\n7. Now suppose that $y \\ne 0$. Then $\\langle y, y \\rangle \u003e 0$ by axiom 3. Put $c = \\langle x, y \\rangle / \\langle y, y \\rangle$ and $z = x - c y$.\n8. By axiom 1, $\\langle z, y \\rangle = \\langle x, y \\rangle - c \\langle y, y \\rangle = 0$. By axiom 2, also $\\langle y, z \\rangle = \\overline{0} = 0$.\n9. Since $x = z + c y$, axiom 1 and step 2 give $\\langle x, x \\rangle = \\langle z, z \\rangle + \\bar c \\langle z, y \\rangle + c \\langle y, z \\rangle + c \\bar c \\langle y, y \\rangle$.\n10. By step 8 the middle terms vanish. A [[complex number times its conjugate|0ZB7J071WQ]] is $c \\bar c = \\abs{c}^2$. So $\\norm{x}^2 = \\norm{z}^2 + \\abs{c}^2 \\norm{y}^2$.\n11. The [[absolute value is multiplicative|0GMV7GP8W0]], so $\\abs{c}^2 = \\abs{\\langle x, y \\rangle}^2 / \\norm{y}^4$. Hence $\\abs{c}^2 \\norm{y}^2 = \\abs{\\langle x, y \\rangle}^2 / \\norm{y}^2$.\n12. Since $\\norm{z}^2 \\ge 0$, steps 10 and 11 give $\\abs{\\langle x, y \\rangle}^2 \\le \\norm{x}^2 \\norm{y}^2$. Equality holds exactly when $\\norm{z} = 0$, that is, when $z = 0$.\n13. Both $\\abs{\\langle x, y \\rangle}$ and $\\norm{x} \\norm{y}$ are $\\ge 0$. Taking square roots in step 12 gives $\\abs{\\langle x, y \\rangle} \\le \\norm{x} \\norm{y}$, with equality exactly when $z = 0$.\n14. If $z = 0$, then $1 \\cdot x - c y = 0$. So $x$ and $y$ are [[linearly dependent|0EVHJFTG9Z]].\n15. Conversely, let $a x + b y = 0$ with $a, b \\in \\mathbb{K}$ not both $0$. If $a = 0$, then $b \\ne 0$ and $y = 0$, which is excluded. So $a \\ne 0$ and $x = d y$ with $d = -b / a$.\n16. Then $z = (d - c) y$, and step 8 gives $(d - c) \\langle y, y \\rangle = \\langle z, y \\rangle = 0$. Since $\\langle y, y \\rangle \\ne 0$, we get $d = c$ and $z = 0$.\n17. By steps 13 to 16, when $y \\ne 0$, equality holds if and only if $x$ and $y$ are linearly dependent. With step 6, this proves the theorem.\n\n## Idea\n\nSubtract from $x$ its projection $c y$ onto the line through $y$. The remainder $z$ is orthogonal to $y$, so a Pythagoras identity gives $\\norm{x}^2 = \\norm{z}^2 + \\abs{c}^2 \\norm{y}^2$. Dropping $\\norm{z}^2$ gives the inequality, and equality means $z = 0$.\n",
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