= Minimizers under strictly increasing transformations
{title2=$\operatorname*{arg\,min}\phi\circ f=\operatorname*{arg\,min}f$}
If $\phi$ is <strictly increasing> on the range of $f$, then $f(x)\le f(y)$ if and only if $\phi(f(x))\le\phi(f(y))$. Quantifying over every $y$ proves that $f$ and $\phi\circ f$ have the same <argmin>, whether or not a minimum is attained. Squaring therefore preserves the <argmin> of a nonnegative objective, while squaring a signed objective need not preserve its <argmin>.
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