<s>[INST] If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$? This question comes in mind while solving another question.

If $abc=1$, does $(a^2+b^2+c^2)^2 \geq a+b+c$ ?

I wonder if this question (if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$) helps?
I wondered if AM-GM could help, but the extra square keeps bothering me while solving it.
Another thought: If $(a^2+b^2+c^2)^2 \geq 1$, then this statement will be true. But how can I prove it?
 [/INST] I'll suppose $a$, $b$, $c>0$ as that always seems to be assumed in these
sorts of problems. Then your inequality is equivalent to
$$(a^2+b^2+c^2)^2\ge abc(a+b+c)=a^2bc+b^2ac+c^2ab.$$
As $bc\le\frac12(b^2+c^2)$ etc. we get
$$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2$$
and that is clearly less than $(a^2+b^2+c^2)^2$.
We can do better:
$$\frac{a^4+b^4}2+2a^2b^2\ge3a^2b^2$$
etc. Therefore
$$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2
\le\frac13(a^2+b^2+c^2)^2.$$
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