I'm new to this community. Do you think my solution is correct? The problem is: $$a^2+b^2+c^2+ab+bc+ca\geq6, a+b+c=3$$
My solution: $$a^2+b^2+c^2+2ab+2bc+2ca\geq3(ab+bc+ca)\rightarrow$$ $$3\geq ab+bc+ca$$
By AM-GM (1): $$a^2+b^2+c^2\geq 2(a+b+c)-3$$ By AM-GM: $$a^2+b^2+c^2+ab+bc+ca\geq2(a+b+c)-3+ab+bc+ca\geq2(a+b+c)-ab+bc+ca+ab+bc+ca=2(a+b+c)=6$$
Assuming $a+b+c=3$ is given, I would start with $(a+b+c)^2=9 \iff a^2+b^2+c^2+ab+bc+ca=9-(ab+bc+ca)$, hence we are asked to prove $$9-(ab+bc+ca)\ge 6$$ $$\iff 3\ge ab+bc+ca$$ for which you have already a valid proof.