Prove for all reals: $a^2+b^2+c^2\geq 2(a+b+c)-3$ Problem:

Prove this for all reals $a,b,c$:
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$

Attempt:
I am trying to work backwards.
$$a^2+b^2+c^2\geq 2(a+b+c)-3$$
$$(a+b+c)^2\geq 2(a+b+c)-3 + 2(ab+bc+ca)$$
$$(a+b+c)^2-2(a+b+c)\geq 2(ab+bc+ca)-3  $$
$$(a+b+c)(a+b+c-2)\geq 2(ab+bc+ca)-3  $$
I am stuck here. And need hints.

it is equivalent to $$(a-1)^2+(b-1)^2+(c-1)^2\geq 0%$$
