$a^2+b^2+c^2=(abc)^2-2\leq 6$. 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!
I found the reversed claim is true, which means the claim is false.
Put $s=x+y+z, x = a^2, y = b^2, z = c^2 \Rightarrow x+y+z = xyz-2 \leq \dfrac{(x+y+z)^3}{27} - 2 \Rightarrow s \leq \dfrac{s^3}{27}-2\Rightarrow s^3-27s-54 \geq 0 \Rightarrow (s-6)(s+3)^2 \geq 0 \Rightarrow s \geq 6.$
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 $6 \le 6$

$a,b,c = \sqrt{2} &gt; 0$