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comment Division by $0$
@JMCF125 then you roll
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revised How can I prove that $xy\leq x^2+y^2$?
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comment How can I prove that $xy\leq x^2+y^2$?
@ronno Not really, because $X := \mathbb{R}^2 \setminus \{x = -y\}$ is dense in $\mathbb{R}^2$, and $[0, +\infty)$ is closed in $\mathbb{R}$, therefore its pullback along $(x, y) \mapsto x^2 - xy + y^2$ must be $\operatorname{cl}X = \mathbb{R}^2$. Of course it is too technical for a precalculus-level question.
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revised How can I prove that $xy\leq x^2+y^2$?
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