# $a^2+b^2+c^2=(abc)^2-2\leq 6$. 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!

• If someone posts a proof, I need a proof that doesn't use calculus. – inequals May 6 '15 at 6:18

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

• This makes me sad. :\ – inequals May 6 '15 at 6:33
• Yes, I knew that as well since you probably are hoping it is true so you can prove the harder inequality. – DeepSea May 6 '15 at 6:34
• Yup! I was actually trying at this one. – inequals May 6 '15 at 6:34
• Need to think of some new trick now. :P – inequals May 6 '15 at 6:35
• you might want to look at some techniques like EV at AOP. – DeepSea May 6 '15 at 6:36

Let $t=a^2+b^2+c^2\geq 0$. We have $$(abc)^2=a^2b^2c^2\leq\frac{(a^2+b^2+c^2)^3}{27}=\frac{t^3}{27}.$$ It follows from $a^2+b^2+c^2=(abc)^2-2$ that $$t\leq\frac{t^3}{27}-2.$$ Therefore $$(t+3)^2(t-6)\geq 0.$$ Since $(t+3)^2\geq 9$, we have $t\geq 6$.

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 $6 \le 6$

$a,b,c = \sqrt{2} > 0$