From a step in a convex analysis proof: $$\large \inf_{v \ge 0}(\frac{(d - v)^2}{v})= \{4d \text{ if } d \le 0, 0 \text{ if } d \ge 0\}$$
Can someone explain why this is true? I can't see why this is the case. I can see why for $d \ge 0$ because $b=d$ gives $0$, but for $d\le0$ I can't figure how this is derived.
I think it should be $-4d$. May be I am making a mistake somewhere which you could fix. So our question is to extremize $f(v)=\frac{(d-v)^2}{v},v\ge 0$. Now $f'(v)=1-\frac{d^2}{v^2}=0$ gives us $v=\pm d$.
Case I: For $d>0$, we must take $v=d$ since it is given that $v\ge 0$. By First derivative test, minimum occurs at $v=d$ given by $f(d)=0$.
Case II: For $d<0$, we must take $v=-d$ since it is given that $v\ge 0$. By First derivative test, minimum occurs at $v=-d$ and and is given by $f(-d)=-4d$