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If $$a+b+c=3abc$$ 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

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    $\begingroup$ $a=b=c=1{{{}}}$? $\endgroup$ – Angina Seng Mar 3 '19 at 10:52
  • $\begingroup$ Any conditions on $a,b,c$, like being positive? $\endgroup$ – Ingix Mar 3 '19 at 10:56
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    $\begingroup$ $ab+bc+ca \geq 3 (abc)^{\frac 2 3}.$ $\endgroup$ – little o Mar 3 '19 at 10:58
  • $\begingroup$ @lngix thanks for reminding me!! $\endgroup$ – Avi Mar 3 '19 at 10:59
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    $\begingroup$ Might not be provable because it's false. $a = b = c =1$ satisfies the antecedent and fails the consequent. $\endgroup$ – kakashi10192020 Mar 3 '19 at 11:01
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Since $${a+b+c\over 3}\geq \sqrt[3]{abc}$$

we get $$abc\geq \sqrt[3]{abc} \implies a^3b^3c^3\geq abc \implies a^2b^2c^2 \geq 1$$

Since $${x+y\over 2}\geq \sqrt{xy} \implies x+y\geq 2\sqrt{xy}$$

so we have $$(a+b)(b+c)(c+a)\geq 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8\sqrt{a^2b^2c^2}\geq 8$$

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  • $\begingroup$ thanks man!! but I am dumb and I got the question wrong,it would be extremely helpful if you can go through it once again mate! $\endgroup$ – Avi Mar 4 '19 at 17:26
  • $\begingroup$ is now any better? $\endgroup$ – Aqua Mar 4 '19 at 17:52
  • $\begingroup$ I guess second step is wrong...since if x*y=1 and x>k this doesn't implies y>k.which you did on second step $\endgroup$ – Avi Mar 5 '19 at 14:34
  • $\begingroup$ What? Do you know what are you asking? $\endgroup$ – Aqua Mar 5 '19 at 14:35
  • $\begingroup$ Maybe you didn't saw that I changed the question(since I got it wrong the first time).I am sorry for any inconvenience. $\endgroup$ – Avi Mar 5 '19 at 14:37
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By AM-GM $$a+b+c=3abc\leq3\left(\frac{a+b+c}{3}\right)^3,$$ which gives $$a+b+c\geq3.$$ Thus, $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)\geq$$ $$\geq\frac{8}{9}(a+b+c)\sqrt{3abc(a+b+c)}=\frac{8}{9}(a+b+c)^2\geq8.$$

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  • $\begingroup$ Thanks man,but since I am dumb I got the question wrong can you please go through it once again? $\endgroup$ – Avi Mar 4 '19 at 17:24

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