Algebraic way to show $112\sqrt{3}n^2-112\sqrt{3}-21n+112$ is positive for $n\ge 1$

Question

As in the title, i am looking for a nice algebraic way to show $112\sqrt{3}n^2-112\sqrt{3}-21n+112$ is positive for $n\ge 1$. Without appealing to derivatives if possible. I am not sure if this possible but I would appreciate any pointers.

Answer 1

The whole thing is equal to: $$112\sqrt{3}n(n-1)+(112\sqrt{3}-21)(n-1)+91\geq 91 >0.$$

Answer 2

$p=112 \sqrt 3 n^2 -112\sqrt 3-21n + 21 +91=112 \sqrt 3 (n^2-1) -21(n-1)+91=(n-1)[(n+1)( 112 \sqrt 3-21)] +91$

$112 \sqrt 3 > 21$

So fo $n=1,.. or..n>1 $ ,.. $p>0$