<s>[INST] $a^2+b^2+c^2=(abc)^2-2\leq 6$. Proof or counter-example needed for $a,b,c\gt 0$ I was working on an inequality proof in which I need to use the following inequality to conclude.

$$\forall~a,b,c\gt 0~,~a^2+b^2+c^2=(abc)^2-2\implies a^2+b^2+c^2\leq 6$$

I can't think of any way to prove this. I only need a hint since I dislike people just giving away the answers. Thanks in advance!
If this claim is false, I'd like a counter-example. Thanks again!
 [/INST] True for $a = b = c$, then $a = \sqrt{\frac{6}{3}} = \sqrt{2}$
$\sqrt{2}^2 + \sqrt{2}^2 + \sqrt{2}^2 = (\sqrt{2} \cdot \sqrt{2} \cdot \sqrt{2})^2 - 2$
$6 = 8-2$ and $6 \le 6$
$a,b,c = \sqrt{2} > 0$
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