Bài 103189

Question

Cho $\begin{cases}a_{1},a_{2},...,a_{n}\geq 0 \\ a_{1}+a_{2}+...+a_{n}=1 \\ n \in Z,n\geq 2\end{cases}$
Chứng minh rằng:
$\left ( 1+\frac{1}{a_{1}} \right )\left ( 1+\frac{1}{a_{2}} \right )...\left ( 1+\frac{1}{a_{n}} \right )\geq \left ( n+1 \right )^{n}$

score 0 · Answer 1

Theo BĐT Cauchy:
$1+\frac{1}{a_{i}}=\frac{1}{a_{i}}\left (a_{i}+a_{1}+a_{2}+...+a_{n} \right )\left ( \forall i=1,2,...,n \right )$
$\geq\frac{1}{a_{i}}\left ( n+1 \right )\sqrt[n+1]{a_{1}a_{2}...a_{n}a_{i}}$
$\Rightarrow 1+\frac{1}{a_{i}}\geq \left ( n+1 \right )\sqrt[n+1]{\frac{a_{1}a_{2}...a_{n}}{a^{n}_{i}}}$
$\Rightarrow \left ( 1+\frac{1}{a_{1}} \right )\left ( 1+\frac{1}{a_{2}} \right )...\left ( 1+\frac{1}{a_{n}} \right )\geq \left ( n+1 \right )^{n} . \sqrt[n+1]{\frac{\left ( a_{1}a_{2}...a_{n}\right )^{n}}{a^{n}_{1}.a^{n}_{2}...a^{n}_{n}}}= \left ( n+1 \right )^{n}$
Dấu "=" xảy ra $\Leftrightarrow a_{1}=a_{2}=...=a_{n}=\frac{1}{n}$

Bài 103189

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Bài 103189

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