Given any integer $n$, define $$ f(t) := (2+(n-2)t^2)^4$$ and $$ g(t) := (n-2)^2t^8 + (n^4-(n-2)^2)t^4.$$
One can easily see that $f(1) = g(1)$.
By doing numerical simulations, I can see that $f(t) > g(t)$ for any real $t > 1$ and any $n \geq 5$. How can we prove this analytically ?
I think it should be $g(t)=(n-2)^2t^8+(n^4-(n-2)^2)t^4$.
If so $$f(t)-g(t)=(t^2-1)((n^4-8n^3+23n^2-28n+12)t^6+(n^4-25n^2+68n-52)t^4-(32n-48)t^2-16)\geq$$ $$\geq(t^2-1)((12n+12)t^2+(68n-52)t^2-(32n-48)t^2-16)\geq$$ $$\geq(t^2-1)(48n-8)>0.$$