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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!

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are $m,n\in\mathbb{N}$? – mathemagician Jan 20 '13 at 15:04
@mathemagician, sorry, I intended $n \in \mathbb{N}$, but not necessarily $m$. – user58997 Jan 20 '13 at 15:15
up vote 1 down vote accepted

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.

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Beautiful solution. Thanks, @Nehsb! – user58997 Jan 20 '13 at 15:12
I would have upvoted your solution, too, but don't have the reputation... – user58997 Jan 20 '13 at 15:27

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