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For positive numbers $a,b,c$ we have $(a+b)(a+c)(b+c)=8abc$. Prove $a=b=c$

I tried expanding the expression. after simplifying we have,

$$a^2b+ab^2+b^2c+ca^2+ac^2+bc^2=6abc$$ But not sure how to continue.

I also noticed that we have,

$$(a+b)(a+c)(b+c)=(2a)(2b)(2c)$$ $$(a+b)+(a+c)+(b+c)=(2a)+(2b)+(2c)$$ But I don't know if it helps.

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2 Answers 2

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By AM-GM, $$a+b \geqslant 2\sqrt{ab}$$ with equality if and only if $a=b$. Multiplying together the three similar inequalities we get $$(a+b)(b+c)(c+a) \geqslant 8abc$$ with equality if and only if $a=b=c$.

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    $\begingroup$ Thanks! I was thinking about AM-GM on all three numbers and missed that we can use it for each two numbers! $\endgroup$
    – Amirali
    Dec 14, 2021 at 14:19
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Continuing from where you left, you have: $b(a-c)^2+a(b-c)^2+ b^2c+ca^2-2abc=0$,which is same as $$b(a-c)^2+a(b-c)^2+c(b-a)^2=0$$ So you now have sum of three non-negative numbers equal to $0$. Can you take it from here?

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