<s>[INST] 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?
 [/INST] Famke's answer is the simplest; however, we can use a variational argument, as well.
For all variations that maintain $ab+bc+ca=2$, we have
$$
(b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
To minimize $a^2+b^2+c^2$, we must have
$$
2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2}
$$
for all variations that satisfy $(1)$.
Orthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need
$$
2a=\lambda(b+c),\quad2b=\lambda(a+c),\quad\text{and}\quad2c=\lambda(a+b)\tag{3}
$$
Summing these up, we get that $\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\sqrt{\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.
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