$(m_1+n_1)(m_2+n_2)(m_3+n_3)\geq8m_1m_2m_3$ $(m_1-n_1)(m_1+n_1)(m_2-n_2)(m_2+n_2)(m_3-n_3) \ge 8m_1m_2m_3(m_1m_2m_3-n_1n_2n_3)$ $(m_1^2-n_1^2)(m_2^2-n_2^2)(m_3^2-n_3^2) \ge 8m_1^2m_2^2m_3^2-8m_1m_2m_3n_1n_2n_3$ From here, maybe we can expand and manipulate?

Let $p=\frac{m_1}{n_1},q=\frac{m_2}{n_2},r=\frac{m_3}{n_3}$, we get $(p-1)(q-1)(r-1)=pqr-1\rightarrow pq+qr+rp=p+q+r$ And we want to prove that $(p+1)(q+1)(r+1)\geq 8pqr$ $\Leftrightarrow 2(\sum_{cyc}{pq})^2(\sum_{cyc}{p})^2+(\sum_{cyc}{ab})^3\geq 7pqr(\sum_{cyc}{p})^3$ which is true.

Let $x=\frac{m_1}{n_1},y=\frac{m_2}{n_2},z=\frac{m_3}{n_3}$, we get $(x-1)(y-1)(z-1)=xyz-1\rightarrow xy+yz+zx=x+y+z$ Let now $p=x+y+z=xy+yz+zx=q$ and $r=xyz$ we want to prove that: $2p^2q^2+p^3r>=7q^3$ which is true because $p^3r>=q^3$ and $p^2>=3q$

P2nisic wrote: which is true because $p^3r>=q^3$ and $p^2>=3q$ Are you sure?

A similar one. We have $(m_1-n_1)(m_2-n_2)(m_3-n_3)=m_1m_2m_3-n_1n_2n_3\iff \sum{m_1m_2n_3}=\sum{n_1n_2m_3}$ By dividing both sides by $n_1n_2n_3$ gives $\sum{\frac{m_1m_2}{n_1n_2}}=\sum{\frac{m_1}{n_1}}$ Let's write $\frac{m_1}{n_1}=x,\frac{m_2}{n_2}=y,\frac{m_3}{n_3}=z$ We have $xy+yz+zx=x+y+z$ and we want to prove that $2(xy+yz+zx)+1\geq 7xyz$ which can be obtained by dividing both sides by $n_1n_2n_3$ in desired ineqaulity.

Let $\frac{m_1}{n_1}=x, \frac{m_2}{n_1}=y, \frac{m_3}{n_3}$. We have $$xy+yz+zx=x+y+z$$. What we want to prove is $$xy+yz+zx+x+y+z+1\ge 7xyz$$. Using $xy+yz+zx=x+y+z$ we can homogenize the inequality into $$2(xy+yz+zx)^2(x+y+z)^2+(xy+yz+zx)^3 \ge 7xyz(x+y+z)^3$$. Expanding and cancelling out terms gives $$2 \sum_{sym} x^4y^2+ 5\sum_{cyc}x^3y^3 \ge 3xyz\sum_{cyc}x^3+2xyz\sum_{sym}x^2y+6x^2y^2z^2$$. Which we can prove with Muirhead or AM-GM Inequality.