<s>[INST] 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:
$a^2+b^2+c^2+ab+bc+ca\ge6$
 [/INST] By squaring $a+b+c=3$ we get
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=9.$$
From the AM-GM inequality 
(or from the fact that $(x-y)^2=x^2+y^2-2xy\ge 0$, i.e. $2xy\le x^2+y^2$)
we have
$$ab+ac+bc \le \frac{a^2+b^2}2+\frac{a^2+c^2}2+\frac{b^2+c^2}2=a^2+b^2+c^2,$$
i.e. $\frac12(a^2+b^2+c^2) \ge \frac12(ab+ac+bc)$, which is equivalent to
$\frac12(a^2+b^2+c^2) - \frac12(ab+ac+bc) \ge0$.
By adding the above equality and inequality together you get
$$\frac32(a^2+b^2+c^2+ab+ac+bc)\ge9,$$
which is equivalent to
$$a^2+b^2+c^2+ab+ac+bc\ge6.$$
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