<s>[INST] Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers 
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers

It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof using that or otherwise, too?
 [/INST] We use AM-GM three times
$$\frac{a^2 +a^2+a^2+a^2 +b^2+c^2}{6} \geq a (abc)^{1/3}$$
$$\frac{b^2 +b^2+b^2+b^2 +a^2+c^2}{6} \geq b (abc)^{1/3}$$
$$\frac{c^2 +c^2+c^2+c^2 +b^2+a^2}{6} \geq c (abc)^{1/3}$$
Summing these inequalities and dividing by 6 gives 
$$a^2+b^2+c^2 \geq (a+b+c)(abc)^{1/3}$$
Now using that $abc=1$, we conclude the desired inequality.
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