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  "body": "\n## Proof\n\n1. The inequality $\\norm{x + y} \\le \\norm{x} + \\norm{y}$ is axiom 1 in the definition of a [[norm|0V2WP9B1MF]].\n2. Write $x = (x - y) + y$. By axiom 1, $\\norm{x} \\le \\norm{x - y} + \\norm{y}$. So $\\norm{x} - \\norm{y} \\le \\norm{x - y}$.\n3. In the same way, $y = (y - x) + x$ gives $\\norm{y} - \\norm{x} \\le \\norm{y - x}$.\n4. Since $y - x = (-1)(x - y)$, axiom 2 gives $\\norm{y - x} = \\abs{-1} \\norm{x - y}$. The [[absolute value|0KW2X0TFBX]] of $-1$ is $1$, so $\\norm{y - x} = \\norm{x - y}$.\n5. Put $t = \\norm{x} - \\norm{y}$, a real number. By steps 2 to 4, both $t \\le \\norm{x - y}$ and $-t \\le \\norm{x - y}$.\n6. The absolute value $\\abs{t}$ equals $t$ or $-t$. So step 5 gives $\\bigl| \\norm{x} - \\norm{y} \\bigr| \\le \\norm{x - y}$.\n\n## Idea\n\nThe first inequality is an axiom of a norm. For the reverse form, apply the axiom to $x = (x - y) + y$ and to $y = (y - x) + x$. Since $\\norm{y - x} = \\norm{x - y}$, the two bounds together control $\\abs{\\norm{x} - \\norm{y}}$.\n",
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