Equality of an Inequality

I encountered the following problem in my textbook-

Let $a, b, c$ be three arbitrary real numbers. Denote $$ x = sqrt{b^2-bc+c^2}, y = sqrt{c^2-ca+a^2}, z = sqrt{a^2-ab+b^2} $$ Prove that $$ xy+yz+zx ge a^2+b^2+c^2 $$

Textbook’s Solution:
Rewrite x,y in the following forms $$ x = sqrt{{3c^2over 4}+left(b-{cover 2}right)^2}, y = sqrt{{3c^2over 4}+left(a-{cover 2}right)^2} $$
According to Cauchy-Schwarz inequality, we conclude $$ xy ge {3c^2over 4}+{1over 4}left( 2b-cright)left(2a-c right) $$
which implies
$$ sum_{cyc}xy ge {3over 4}sum_{cyc}c^2 +{1over 4}sum_{cyc}left( 2b-cright)left(2a-c right) = sum_{cyc}a^2. $$
My Approach:
From the given values of $x$ and $y$–
$$ xy = sqrt{left( c^2-bc+b^2right) left(c^2-ca+a^2 right)} ge c^2+csqrt{ab}+ab $$
So, $$xy+yz+zx ge sum_{cyc}a^2+sum_{cyc}asqrt{bc}+sum_{cyc}bc$$
And it rests to prove that-
$$ sum_{cyc}a^2+sum_{cyc}asqrt{bc}+sum_{cyc}bc ge sum_{cyc}a^2 $$
$$ sum_{cyc}asqrt{bc}+sum_{cyc}bc ge 0 $$
Now my question is that how to prove this? Or are we done?
And also if both approachs are correct, what is the equality case?
The equality case is not given in the textbook so I ask.
Thanks!