Balkan Mathematical Olympiad Shortlist

Divide both sides of the given equality by $m_1{m}_2{m}_3$ and set $a=\frac{n_1}{m_1}$, $b=\frac{n_2}{m_2}$ and $c=\frac{n_3}{m_3}$. The equality $(m_1-n_1)(m_2-n_2)(m_3-n_3)=m_1{m}_2{m}_3-n_1{n}_2{n}_3$ becomes $$(1-a)(1-b)(1-c)=1-abc⟺a+b+c=ab+bc+ca$$ and we have to show that $$(a+1)(b+1)(c+1)\ge 8⟺a+b+c+ab+ba+ca+abc\ge 7.(1)$$ Let $a+b+c=ab+bc+ca=t$. Since $(a+b+c{)}^2\ge 3(ab+bc+ca)$, we have $t^2\ge 3t$, i.e. $t\ge 3$. Furthermore $$a^3+b^3+c^3\ge \frac{(a^2+b^2+c^2{)}^2}{a+b+c}=\frac{(t^2-2t{)}^2}{t}=t(t-2{)}^2$$ and $$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=t(t^2-3t)+3abc$$ imply $3abc\ge t(t-2{)}^2-t^2(t-3)=4t-t^2$. Since (1) is equivalent to $2t+abc\ge 7$, which is true for $t\ge \frac{3}{2}$, it suffices to show that $2t+\frac{4t-t^2}{3}\ge 7$ for $t\in [3,\frac{3}{2}]$. The latter is equivalent to $(t-3)(t-7)\le 0$, which is true for $t\in [3,7]$.