Question

Prove:
(x+1)(x+2)(x+3)(x+4). . .(x+2012)(x+2013) <= (x+1007)^2013

Answer 1

the exponent seems clear to mean, the average between (x+1) and (x+2013) is x = 1007

Likewise the average between 1 and 2013 is 1007

Answer 2

Average of summands between brackets

\[ n = \frac{2x+2014}{2} = \frac{2(x+1007)}{2}=x+1007 \]

Answer 3

Maybe this can be proved if it's split up as an arithmetic series, the rest will just be a multiplication

\[ \Pi_0^{2013} \]

Answer 4

*bookmark

Answer 5

I think the trick is to pair the factors like so \[(x+1)(x+2013)(x+2)(x+2012)...(x+1006)(x+1008)(x+1007)\]We need to verify that\[(x+a)(x+2014-a)\le(x+1007)^2\]When \(1\le a \le1006\). Expand that out, and we get, \[x^2+2014x+2014a-a^2 \overset{\bf{}?}\leq x^2+2014x+1014049\]Simplify a bit, and we get to \[a^2-2014a+1014049\overset{\bf{?}}\ge0\]Now here's the slightly tricky part. Notice that this factors as\[(a-1007)^2\overset{\bf{?}}\ge0\implies |a-1007|\overset{\bf{?}}\ge0\]Notice that according to our assumption, \(1\le a\le1006\), this means that we're done.

Answer 6

Since every pair \((x+a)(x+2014-a)\le(x+1007)^2\), and we have 1006 pairs, that means we have that \[(x+1)(x+2013)(x+2)(x+2012)...)(x+1006)(x+2008)\le(x+1007)^{2012}\]Tack on an extra \((x+1007)\leq(x+1007)\), and we are done.

Answer 7

Thanks All...