How can I establish this inequality? I'm trying to show that for $m\geq 1$ and $n>2$, we have the following inequality:

$4(m-1)^{n} + 2nm^{n-1} \geq (m-2)^{n} + 3 m^{n}$

Any ideas?
Thanks for the help!
 A: Let $u=m-1.$
Then, we want to show $$4u^n+2n(u+1)^{n-1} \ge (u-1)^n+3(u+1)^n$$
As polynomials, the coefficient of $u^i$ on the left for $i<n$ is $2n{n-1\choose i},$ while on the right it is $2{n\choose i}$ for $n-i$ odd and $4{n\choose i}$ for $n-i$ even. This is by the binomial theorem. The coefficient of $u^n$ is 4 on both the left and the right side. 
Note that $n{n-1\choose i} = (n-i){n\choose i}.$ I claim that the coefficient of $u^i$ on the left is greater than the coefficient of it on the right for $i \le n-2$, and is equal to it for $i = n-1, n$. This will solve the problem as $u$ must be nonnegative.
Indeed, for $i = n-1$, $2n{n-1\choose n-1} = 2{n \choose n-1}.$ For $i=n$, both coefficients are 4. And for $i \le n-2$, $2n{n-1\choose i} = 2(n-i){n \choose i}\ge 4{n\choose i},$ the largest possible value of the coefficient on the right.
If we are allowing $m,n$ to not be natural numbers, then this still works if $m > 2$, just use the extended binomial theorem instead of the binomial theorem. And if $m < 2$, then $(m-2)^n$ isn't defined for $n$ not a natural number, and the argument works if $n$ is.
