# How to prove the follwing inequality $(a+b)(b+c)(c+a) \ge 8$

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

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

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

• 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! – Avi Mar 4 '19 at 17:26
• is now any better? – Aqua Mar 4 '19 at 17:52
• 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 – Avi Mar 5 '19 at 14:34
• What? Do you know what are you asking? – Aqua Mar 5 '19 at 14:35
• Maybe you didn't saw that I changed the question(since I got it wrong the first time).I am sorry for any inconvenience. – Avi Mar 5 '19 at 14:37

By AM-GM $$a+b+c=3abc\leq3\left(\frac{a+b+c}{3}\right)^3,$$ which gives $$a+b+c\geq3.$$ Thus, $$(a+b)(a+c)(b+c)\geq\frac{8}{9}(a+b+c)(ab+ac+bc)\geq$$ $$\geq\frac{8}{9}(a+b+c)\sqrt{3abc(a+b+c)}=\frac{8}{9}(a+b+c)^2\geq8.$$

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