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:
$a^2+b^2+c^2+ab+bc+ca\ge6$

$$ a^2 + b^2 + c^2 + ab + bc + ac = (a+b+c)^2 - (ab + bc + ac) = 9 -  (ab + bc + ac)$$
Now it remains to show that max value of $(ab + bc + ac)$ is $3$.  For that, we know the AM-GM equality ( for $a,b, c >0$ ) that $3(a^2 + b^2 + c^2) \geq (a+ b + c)^2 \geq 3(ab +bc +ac)$. From the last two part we have $(a+ b + c)^2 \geq 3(ab +bc +ac) \implies 9 \geq 3 (ab +bc +ac) \implies 3 \geq ab +bc +ac$
Hence we have 
$$ a^2 + b^2 + c^2 + ab + bc + ac = 9 -  (ab + bc + ac) \geq 9 - 3 = 6$$
