3

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?

Sujaan Kunalan
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Richard
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    (1) The third and fourth lines are redundant in view of the second line. (2) $a+b+c\ge 3$ does not imply that each of $a,b,c\ge 1$. – vadim123 Jul 30 '13 at 17:47
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    Probably we want to ask that the numbers be positive, else $a=b=-3$, $c=1$ gives trouble. – André Nicolas Jul 30 '13 at 17:52

1 Answers1

8

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

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

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

Calvin Lin
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