Proving the inequality $a^2+b^2+c^2+ab+bc+ca\ge6$ Given that $a$, $b$, $c$ are non-negative real numbers such that $a+b+c=3$, how can we prove that:
$a^2+b^2+c^2+ab+bc+ca\ge6$

By the Cauchy-Schwarz inequality,
$$6=1*(a+b)+1*(b+c)+1*(a+c)\leq\sqrt{1^2+1^2+1^2}\sqrt{(a+b)^2+(b+c)^2+(a+c)^2}$$
In other words
$$(a+b)^2+(b+c)^2+(a+c)^2\geq 12$$
which is the same as the desired inequality.
