Let $x\geq 1, n\in N$ Show that $$ \prod_{i=1}^n(x+n-i)\geq x^{n+1} $$

I can see that the product is greater or equal to $x^n$, but I cannot understand where power $n+1$ is coming from.


closed as off-topic by Martin R, Paul Frost, воитель, Javi, blub Aug 13 at 20:04

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  • 1
    $\begingroup$ Doesn't the inequality fails for $n=1$? Are you sure the low index is $i=0$? $\endgroup$ – Pspl Aug 13 at 11:12
  • 3
    $\begingroup$ The LHS is a polynomial of degree $n$ in $x$, so this is wrong for any $n$ and sufficiently large $x$. $\endgroup$ – Martin R Aug 13 at 11:13

I'm pretty sure you wanted to write $\prod_{i=0}^n(x+n-i)\geq x^{n+1}$, for all $x\geq 1$, $n\in \mathbb{N}$. Please note that your inequality fails if $n=1$. If that's the case, you can use induction:

If $n=1$ then

$$\begin{align*} \prod_{i=0}^n(x+n-i)&=\prod_{i=0}^1(x+1-i)\\ &=(x+1)x\\ &=x^2+x\\ &\geq x^2 \;\text{(because}\;x\geq 1 \text{)} \end{align*}$$

Suppose now there's a $p\in \mathbb{N}$ so $\prod_{i=0}^p(x+p-i)\geq x^{p+1}$. So, if $n=p+1$, then

$$\begin{align*} \prod_{i=0}^n(x+n-i)&=\prod_{i=0}^{p+1}[x+(p+1)-i]\\ &=x\prod_{i=0}^p[x+(p+1)-i]\\ &\geq x\prod_{i=0}^p(x+p-i) \;\text{(because}\;x+(p+1)-i\geq x+p-i \text{)}\\ &\geq xx^{p+1} \;\text{(because of the hypotheses)}\\ &=x^{(p+1)+1}\\ &=x^{n+1}\\ \end{align*}$$

So, $\;\prod_{i=0}^n(x+n-i)\geq x^{n+1}$, for all $x\geq 1$, $n\in \mathbb{N}\;\;\;\;QED$

  • $\begingroup$ This looks too complicated. Isn't it clear that $x+n-i \geq x$ for all $i$? $\endgroup$ – Kavi Rama Murthy Aug 13 at 12:10
  • $\begingroup$ @Kavi Rama Murthy, non the less is a valid method, right? But you're right. It is pretty clear that $x+n-i\geq x$. $\endgroup$ – Pspl Aug 13 at 12:15

If the product starts with $i=0$ this is trivial. (If it starts with $i=1$ it is false). Just note that $x+n-i \geq x$ for each $i$ so the product is $\geq x^{n+1}$.


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