<s>[INST] Prove that if $a b c=1$, $a^2+b^2+c^2\ge 3$. (for $a,b,c\in\mathbb{R}$) I can do this problem using calculus minimization techniques, using Lagrange multipliers to find the equations $a^2=b^2=c^2$, so with $abc=1$, $a$, $b$, and $c$, are either $-1$ or $1$ (making sure you don't end up with $abc$ negative). So if the minimum distances are at those points, then $a^2+b^2+c^2=3$ is the closest distance, so all other solutions are further away.
I just pieced together (while writing this) that there's another solution: The arithmetic-geometric inequality gives $3(abc)^{1/3}\le a+b+c$, and by the inequality of this question, we have $9(abc)^{2/3}\le (a+b+c)^2\le 3(a^2+b^2+c^2)$, so since $abc=1$, we have $3\le a^2+b^2+c^2$.
But, is it possible to prove this using other inequality tricks? It's just, the inequality on the link I gave, I had no idea about, so it's uncomfortable to use it.
 [/INST] By AM/GM, we have 
$$\frac{a^2+b^2+c^2}{3}\ge (a^2b^2c^2)^{1/3}=1.$$ 
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