Protip: don't link images straight from Wolfram, they clear the image out so it's broken on TSR after a few minutes.
Anyways, I presume you've got to
(x2+1)2+(y2−1)2+1, now you want to make this expression as small as possible, since you're adding together three terms your best bet is to make each
term as small as possible.
Let's try making the "
1" as small as possible: well it's a constant. So it's stuck as 1.
Let's try making (
y2−1)2 as small as possible. Well it's essentially
(something)2 - so the smallest it can ever get is by making something=0 (i.e:
y=1) to get
(12−1)2=02=0.
It won't get any smaller than that. (something)^2 is always positive, your only hope to make it small is by making it 0.
Now we'll want the same tactic to work with
(x2+1)2. Here, since
x2 is a squared term, the smallest you can make it is 0. So the inside becomes (0^2 + 1)^2 = 1^2 = 1. That's the smallest you can make it.
So all in all, the smallest your entire expression can get is 1 + 0 + 1 = 2.