If $(a+1)(b+1)(c+1)=8,$ then prove $abc \le 1$? My proof:
$\sqrt[3]{(a+1)(b+1)(c+1)} = 2 \le \frac{a+b+c}{3} + 1 $according to AM-GM.
Thus, $1 \le \frac {a+b+c}{3}$.
Also, $ \sqrt[3]{abc} \le \frac{a+b+c}{3}$.
Then, either $\frac {a+b+c}{3} \le 1$ or $1 \le \frac {a+b+c}{3}$.
Suppose the latter equation is true. Then $a+b+c \ge 3$ meaning $a,b,c \ge 1$. If that is the case, then $(a+1)(b+1)(c+1) \ge 8$. Therefore, this does not satisfy the original equation so $1 \le \frac {a+b+c}{3}$ must be true.
Is this a sufficient proof? If not, can someone solve this question?

Your proof doesn't work, as pointed out by vadim

Hint: Apply AM-GM directly to each term, we get that

$$ 8 = (a+1)(b+1)(c+1) \geq 2\sqrt{a} 2 \sqrt{b} 2 \sqrt{c} $$

