# Prove that $a + b + c + 2 >= abc$ [closed]

The product of any pair of variables of three positive numbers $$a, b$$ and $$c$$ doesn’t exceed $$4$$. Prove that $$a + b + c + 2 \geq abc$$.

## closed as off-topic by Peter, Arnaud D., Nosrati, José Carlos Santos, Peter ForemanJul 1 at 10:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Peter, Arnaud D., Nosrati, José Carlos Santos, Peter Foreman
If this question can be reworded to fit the rules in the help center, please edit the question.

• It would be interesting which reason hides behind the reopening-vote. This question shows neither context nor an effort. – Peter Jul 1 at 11:06

$$ab\leq 4, bc\leq 4, ca\leq 4$$ so $$a^2b^2c^2\leq 64$$, $$abc\leq 8$$
$$a+b+c+2 = abc(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ba}+\frac{2}{abc})\geq abc(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{2}{8})=abc$$