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

Question

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?

Answer 1

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

Answer 2

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

Answer 3

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 $$a^2bc+b^2ac+c^2ab\le a^2b^2+a^2c^2+b^2c^2 \le\frac13(a^2+b^2+c^2)^2.$$

Answer 4

By Rearrangement inequality we have that

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