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If $$(a+b+c)abc=3$$ and $$a,b,c > 0$$ prove that $$(a+b)(b+c)(c+a)\geq 8$$

I can fairly easily prove that $(a+b)(b+c)(c+a)\geq8abc$, but then I get stuck.....since then I cannot move forward

If I was to prove that $abc\geq1$ this would have been easy but I am stuck, please help me.

Any help is appreciated, thanks.

edit:I am incredibly sorry that I remembered the question incorrectly

Avi
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    Actually, it is $abc \le1$ which you can see as follows: $3 = abc (a+b+c) \ge 3 (abc)^{4/3}$ (by AM-GM) which is $abc \le 1$. – Andreas Mar 05 '19 at 15:24
  • ya I know $abc\leq1$ ,I was saying that if it was $abc\geq1$ this would have been an easy problem,but now I am stuck. – Avi Mar 05 '19 at 15:47
  • change of variable $a^2bc$ and others respectively may help – Bijayan Ray Mar 05 '19 at 16:04
  • @Bijayan Ray intresting ,can you elaborate?I don't see the profit,except that our if statementreduces to a'+b'+c'. – Avi Mar 05 '19 at 16:39

1 Answers1

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Note: $3=abc(a+b+c)\ge 3(abc)^{4/3}$ (by AM-GM) hence $ abc\le 1$.

Using this we have, using AM-GM again: $$(a+b)(a+c)(b+c) = (a+b+c)(ab+ac+bc) - abc\\ \ge 3 (a+b+c)\sqrt[3]{(abc)^2} - 1\\ = 9 (abc)^{-1/3} - 1\\ \ge 9-1 = 8$$

Andreas
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  • Thanks!!!This solution is actually really mind blowingly simple and clever at the same time!!!that second step(introduction of inequality) and also the third step is actually really elegant! – Avi Mar 05 '19 at 16:33
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    If you find the solution appropriate I'd be happy if you actually accept it. – Andreas Mar 11 '19 at 22:00
  • Is that the tick mark thingy??sorry I didn't knew about that – Avi Mar 14 '19 at 10:34
  • Yes that's it. Thanks for accepting! – Andreas Mar 14 '19 at 12:43