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

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

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  • 2
    $\begingroup$ It would be interesting which reason hides behind the reopening-vote. This question shows neither context nor an effort. $\endgroup$ – Peter Jul 1 at 11:06
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$ab\leq 4, bc\leq 4, ca\leq 4$ so $a^2b^2c^2\leq 64$, $abc\leq 8$

$a+b+c+2 = abc(\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ba}+\frac{2}{abc})\geq abc(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{2}{8})=abc$

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  • $\begingroup$ Nice solution! +1 $\endgroup$ – Michael Rozenberg Jul 1 at 11:04
  • $\begingroup$ Thank you very much! I’ve been struggling with this task fo days! $\endgroup$ – Foorgy Infifcio Aug 11 at 11:52

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