Finding the minimum value of $a^2+b^2+c^2$ 
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$. 

I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me  $a+b+c$, which is unknown. How can I solve this?

You can solve this problem by your starting step.
Indeed, by C-S
$$3(a^2+b^2+c^2)\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$
Thus,
$$3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or
$$a^2+b^2+c^2\geq ab+ac+bc$$ and since $ab+ac+bc\geq2,$ we obtain:
$$a^2+b^2+c^2\geq2.$$
The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$, 
which says that $2$ is a minimal value.
