About the equality in Cauchy-Schwarz

$\left(\displaystyle \sum_{i=1}^n a_i^2\right)\left( \displaystyle \sum_{i=1}^n b_i^2\right)\ge \left( \displaystyle \sum_{i=1}^n a_ib_i\right)^2.$

Equality condition in Cauchy-Schwarz is $\frac{a_i}{b_i}=k$ for all i but what happens when ${b_i} = 0$. When the denominator is zero it should be undefined, but zero may be an equality solution.

E.g. $a^2 + b^2 + c^2 \geq ab + bc + ca$

By multiplying by 2 on both sides and a little rearrangement we get

$(a-b)^2 +(b-c)^2 + (c-a)^2 \geq 0$ Which is true for all reals a,b,c and also equality holds when $a=b=c$ even when $a=b=c=0$

We can also solve this by Cauchy-Schwarz :-

$(a^2 + b^2 + c^2)(b^2 + c^2 + a^2) \geq (ab + bc + ca)^2$

Or, $a^2 + b^2 + c^2 \geq ab + bc + ca$

And equality holds when,

$\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a} = 1$

Therfore, we get, $a=b=c$ but $a=b=c=0$ will make the expression undefined but is also a valid solution.

Please help!!!