Proving $a^2 + b^2 + c^2 \geqslant ab + bc + ca$ 
Let $a, b, c \in \mathbb{R}$. Show that $a^2 + b^2 + c^2 \geqslant ab + bc + ca$.

My reasoning went as follows and I would like to know if it's correct.
$a^2 + b^2 + c^2 \geqslant ab + bc + ca$
$\Leftrightarrow a^2 + b^2 + c^2 -ab - bc - ca \geqslant 0$
$\Leftrightarrow 2a^2 + 2b^2 + 2c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow a^2 + a^2 + b^2 + b^2 + c^2 + c^2 -2ab - 2bc - 2ca \geqslant 0$
$\Leftrightarrow (a^2 -2ab + b^2) + (b^2 - 2bc + c^2)+ (c^2 - 2ca + a^2) \geqslant 0$
$\Leftrightarrow (a -b)^2 + (b-c)^2 + (c-a)^2 \geqslant 0$.
And the last inequality holds since squares are non-negative.
I've seen couple of different ways of doing this and they seem to divide by $2$ and I didn't really understand it so instead can I just use the fact that $2a^2 = a^2+a^2$?

Add up the inequalities below 
$$a^2 + b^2 \ge 2ab, \>\>\>\>\>b^2 + c^2 \ge 2bc, \>\>\>\>\>c^2 + a^2 \ge 2ca$$
to arrive at,
$$a^2 + b^2 + c^2 \geqslant ab + bc + ca$$
