Vtyjkyuk

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 $a^2+b^2+c^2ge a+b+c$) 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?

Yes, that question helps a lot! Note that by AM-GM inequality $a^2+b^2+c^2ge 3(a^2b^2c^2)^1/3=3$. Since $x^2geq x$ for all $xgeq 1$, it follows that
$$(a^2+b^2+c^2)^2geq a^2+b^2+c^2geq a+b+c.$$

Hint: use Lagrange multipliers method with $$f(a,b,c)=(a^2+b^2+c^2)^2+a+b+c\g(a,b,c)=abc=1\nabla f=lambdanabla g$$

By Rearrangement inequality we have that

$$(a^2+b^2+c^2)^2ge a^4+b^4+c^4ge a^2bc+b^2ac+c^2ab=abc(a+b+c)=a+b+c$$

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
$$(a^2+b^2+c^2)^2ge abc(a+b+c)=a^2bc+b^2ac+c^2ab.$$

As $bclefrac12(b^2+c^2)$ etc. we get
$$a^2bc+b^2ac+c^2able a^2b^2+a^2c^2+b^2c^2$$
and that is clearly less than $(a^2+b^2+c^2)^2$.

We can do better:
$$fraca^4+b^42+2a^2b^2ge3a^2b^2$$
etc. Therefore
$$a^2bc+b^2ac+c^2able a^2b^2+a^2c^2+b^2c^2
lefrac13(a^2+b^2+c^2)^2.$$