# Let $x\geq 1$, $n\in N$, show that $\prod_{i=1}^n(x+n-i)\geq x^{n+1}$ [closed]

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, blubAug 13 at 20:04

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• Doesn't the inequality fails for $n=1$? Are you sure the low index is $i=0$? – Pspl Aug 13 at 11:12
• The LHS is a polynomial of degree $n$ in $x$, so this is wrong for any $n$ and sufficiently large $x$. – 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 $$x\geq 1$$, $$n\in \mathbb{N}\;\;\;\;QED$$

• This looks too complicated. Isn't it clear that $x+n-i \geq x$ for all $i$? – Kavi Rama Murthy Aug 13 at 12:10
• @Kavi Rama Murthy, non the less is a valid method, right? But you're right. It is pretty clear that $x+n-i\geq x$. – 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 $$\geq x^{n+1}$$.