# If $(a+b)(a+c)(b+c)=8abc$ prove $a=b=c$

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.

• Have you tried AM-GM inequality...? Dec 14, 2021 at 14:12
• Dec 14, 2021 at 14:27

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 $$a=b=c$$.
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?