Why minimize squared L2 norm and not only the L2 norm?

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I'm studying Inverse Problems and usually, they minimize the squared of the L2 norm($L_2, L_0, L_ \infty$), why don't minimize only the norm? if the goal is to have a measure of the distance between 2 parameters?

Why they minimize this: $\left \| m - m_{ref} \right \|_2^2$

instead of:

$\left \| m - m_{ref} \right \|_2 $

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Squaring $\cdot^2\colon [0,\infty)\to [0,\infty)$ is strictly increasing function so minimum will be achieved at the same point: if $f\colon X\to [0,\infty)$, then $$f(m) \leq f(x), \forall x\in X \iff f(m)^2 \leq f(x)^2, \forall x\in X.$$ On the other hand, the square of the $2$-norm is nicer to handle since it gets rid of the square root in the definition of the $2$-norm.