Minimizers under strictly increasing transformations ($\mathrm{argmin}\varphi \circ f=\mathrm{argmin}f$)

If $\varphi$ is strictly increasing on the range of $f$, then $f(x)\le f(y)$ if and only if $\varphi (f(x))\le \varphi (f(y))$. Quantifying over every $y$ proves that $f$ and $\varphi \circ f$ 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.