Bài 104165

Question

Cho $n \in N $\$\left\{ \begin{array}{l} 0,1 \end{array} \right.\left. \right \},x_{1},x_{2},...,x_{n} \geq 0$. Chứng minh các BDT sau:
$a)\sqrt[n]{(n+1)!}\geq 1+\sqrt[n]{n!}$
$b)(1+\frac{x_{1}}{nx_{2}})(1+\frac{x_{2}}{nx_{3}})...(1+\frac{x_{n}}{nx_{1}}) \geq (1+\frac{1}{n})^{n}$

score 2 · Answer 1

a)Áp dụng BĐT Cauchy:
$\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n+1}\geq n\sqrt[n]{\frac{1}{(n+1)!}} (1)$
$\frac{1}{2}+\frac{2}{3}+...+\frac{n}{n+1}\geq n\sqrt[n]{\frac{n!}{(n+1)!}} (2)$
$(1)+(2)$ theo từng vế có:
$n\geq n\sqrt[n]{\frac{1}{(n+1)!}} +n\sqrt[n]{\frac{n!}{(n+1)!}}$
$\Rightarrow \sqrt[n]{(n+1)!}\geq 1+\sqrt[n]{n!}$
Dấu "=" xảy ra $\Leftrightarrow n=1$
b)Áp dụng BĐT Cauchy:
$1+\frac{a}{nb}=\underbrace {\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}}_{n số}+\frac{a}{nb}\geq (n+1)\sqrt[n+1]{\frac{a}{n^{n+1}b}}$
(với $a,b>0$)
$\Rightarrow 1+\frac{a}{nb}\geq \frac{n+1}{n}\sqrt[n+1]{\frac{a}{b}} $
Vì vậy suy ra:
$(1+\frac{x_{1}}{nx_{2}})(1+\frac{x_{2}}{nx_{3}})...(1+\frac{x_{n}}{nx_{1}}) \geq (\frac{n+1}{n})^{n}.\sqrt[n+1]{\frac{x_{1}}{x_{2}}.\frac{x_{2}}{x_{3}}....\frac{x_{n}}{x_{1}}}=(1+\frac{1}{n})^{n} \Rightarrow $ (ĐPCM)

Bài 104165

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Bài 104165

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