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

If $x,y,z$ are nonnegative reals, then
$x^2+y^2+z^2\ge xy+yz+zx$ (with equality iff $x=y=z$), hence
$3(x^2+y^2+z^2)\ge x(x+y+z)+y(y+z+x)+z(z+x+y) = (x+y+z)^2$
(with equality iff $x=y=z$).
Letting $x=a+b, y=b+c, z=a+c$, we find $x+y+z=6$ and
$3(a+b)^2+3(b+c)^2+3(c+a)^2 \ge 36$.
Note that $(a+b)^2+(b+c)^2+(c+a)^2= 2(a^2+b^2+c^2+ab+bc+ca)$ so that we actually showed
$$ a^2+b^2+c^2+ab+bc+ca\ge 6$$
with equality iff $a=b=c$.
