proving an inequality involving three terms The inequality is as follows:
$a^2+b^2+c^2+3\ge2(a+b+c)$
I was thinking about using the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
But I can't get anywhere. I don't have any idea at the moment.

From given equation we have,
$a^2 + b^2 + c^2 − 2a − 2b − 2c + 3 \ge 0$
$(a^2 - 2a +1) + (b^2 - 2b + 1) + (c^2 − 2c + 1) \ge 0$
$(a−1)^2 + (b−1)^2 + (c−1)^2 \ge 0$
