If $(a+b)(a+c)(b+c)=8abc$ prove $a=b=c$
Solution 1:
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 $a=b=c$.
Solution 2:
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?