Is $(1+\alpha)^n \le (1+\alpha)^n1+ (1+\alpha)^n2$ where $\alpha<1$ and $n1+n2=n$? - MathOverflow [closed]

Is $(1+\alpha)^n \le (1+\alpha)^n1+ (1+\alpha)^n2$ where $\alpha<1$ and $n1+n2=n$? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-20T03:41:05Z http://mathoverflow.net/feeds/question/70088 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/70088/is-1-alphan-le-1-alphan1-1-alphan2-where-alpha1-and-n1n2n Is $(1+\alpha)^n \le (1+\alpha)^n1+ (1+\alpha)^n2$ where $\alpha<1$ and $n1+n2=n$? rimpi 2011-07-12T04:39:18Z 2011-07-12T04:39:18Z

<p>Is $(1+\alpha)^n \le (1+\alpha)^{n1}+ (1+\alpha)^{n2}$ where $\alpha<1$ and $n1+n2=n$?</p>