Prove inequality of binomial theorem $\frac1{n^i}\binom{n}i \leq\frac1{m^i}\binom{m}i $.

Question

I have to prove the following inequality:

$$\forall n,m \in \mathbb{N},i\in\mathbb{N_0}:n<m \implies \frac1{n^i}\binom{n}i \leq\frac1{m^i}\binom{m}i $$

I used this fact to prove an other inequality, without proving this one first, but now I am stuck. Is induction the way to do this?

Answer 1

Your inequality is equivalent to

$$m^i\binom{n}i\le n^i\binom{m}i\;,$$

which can be written

$$\frac{m^in^{\underline i}}{i!}\le\frac{n^im^{\underline i}}{i!}$$

and hence is equivalent to

$$m^in^{\underline i}\le n^im^{\underline i}\;.$$

Here $x^{\underline k}=x(x-1)\ldots(x-k+1)$ is a falling factorial. This is clearly true for $i>n$, since the lefthand side is then $0$, and the righthand side is always non-negative. For $i\le n$ it is equivalent to

$$\left(\frac{m}n\right)^i\le\prod_{k=0}^{i-1}\frac{m-k}{n-k}\;,\tag{1}$$

which is clearly true for $i=0$, since both sides are $1$. Finally, it’s easily verified that

$$\frac{m-k}{n-k}\ge\frac{m}n$$

for $0\le k<n$, which establishes $(1)$ for $1\le i\le n$.

Answer 2

We have equality at $i=0$ (both one) and for $i>m$ (both zero). For $n<i\leq m$, the left hand side is zero while the right hand side is positive.

Assuming $0<i\leq n$.

Starting with $m>n$ for any $1\leq r\leq i$ we have $$ \begin{align*} mr&>nr \\ \Rightarrow mn-nr&>mn-mr \\ \Rightarrow \frac{m-r}{m}&>\frac{n-r}{n}. \end{align*} $$ Therefore $$ \begin{align*} \frac{m}{m}\cdot \frac{m-1}{m}\cdot\frac{m-2}{m}\cdots\frac{m-i+1}{m}&>\frac{n}{n}\cdot \frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-i+1}{n} \\\Rightarrow \frac{1}{m^i}\frac{m(m-1)(m-2)\cdots(m-i+1)}{i!}&>\frac{1}{n^i}\frac{n(n-1)(n-2)\cdots(n-i+1)}{i!} \\ \Rightarrow \frac{1}{m^i}{m\choose i}&> \frac{1}{n^i}{n\choose i}\qquad\bullet \end{align*} $$

Answer 3

We have:

$$\frac{m^i}{n^i}\leq\frac{\dbinom{m}i}{\dbinom{n}i}$$

$$\frac{m^i}{n^i}\leq\frac{\dfrac{m!}{i!(m-i)!}}{\dfrac{n!}{i!(n-i)!}}$$

$$\frac{m^i}{n^i}\leq\frac{m!i!(n-i)!}{n!i!(m-i)!}$$

$$\frac{m^i}{n^i}\leq\frac{m!(n-i)!}{n!(m-i)!}$$

$$\frac{m^i}{n^i}\leq\frac{m(m-1)\dots(m-i+1)}{n(n-1)\dots(n-i+1)}$$

So need to prove $\dfrac{m}{n}\le\dfrac{m-1}{n-1}$, or $m(n-1)\le n(m-1)$, or $-m\le -n$, which is true.