Đề bài: Chứng minh rằng: $(\frac{2^{n+1}-1}{n^{2}+2n+1})^{n+1}\geq \frac{C^{0}_{n}C^{1}_{n}C^{2}_{n}…C^{n}_{n}}{(n+1)!},\forall n\in N^{*}$

Lời giải

Đề bài:
Chứng minh rằng: $(\frac{2^{n+1}-1}{n^{2}+2n+1})^{n+1}\geq \frac{C^{0}_{n}C^{1}_{n}C^{2}_{n}…C^{n}_{n}}{(n+1)!},\forall n\in N^{*}$
Lời giải

Ta có: $(1+x)^{n}=\sum\limits_{k=0}^n C^{k}_{n}x^{k}$
$\Rightarrow \int\limits_{0}^{1}(1+x)^{n}dx=\int\limits_{0}^{1}\sum\limits_{k=0}^n C^{k}_{n}x^{k}dx$
$\Rightarrow \frac{2^{n+1}-1}{n+1}=\sum\limits_{k=0}^n\frac{C^{k}_{n}}{k+1}$
Theo BĐT Cauchy:
$\frac{2^{n+1}-1}{n+1}=C^{0}+\frac{1}{2}C^{1}_{n}+\frac{1}{3}C^{2}_{n}+…+\frac{1}{n+1}C^{n}_{n}\geq (n+1)\sqrt[n+1]{ \frac{C^{0}_{n}C^{1}_{n}C^{2}_{n}…C^{n}_{n}}{(n+1)!}}$
$\Rightarrow [\frac{2^{n+1}-1}{(n+1)^{2}}]^{n+1}\geq \frac{C^{0}_{n}C^{1}_{n}C^{2}_{n}…C^{n}_{n}}{(n+1)!}$
$\Rightarrow(\frac{2^{n+1}-1}{n^{2}+2n+1})^{n+1}\geq \frac{C^{0}_{n}C^{1}_{n}C^{2}_{n}…C^{n}_{n}}{(n+1)!}$
$\Rightarrow $ (ĐPCM)

=========
Chuyên mục: Bất đẳng thức Côsi