Bài 104245

Question

Cho $n,p \in Z,n,p \geq 2,a_{ij} >0, \forall i,j \in Z$
$(1 \leq i \leq n, 1 \leq j \leq p)$
Chứng minh rằng:
$\sqrt[n]{(a_{11}+a_{12}+...+a_{1p})(a_{21}+a_{22}+...+a_{2p})...(a_{n1}+a_{n2}+...+a_{np})}$
$\geq \sqrt[n]{a_{11}+a_{21}+...+a_{n1}}+\sqrt[n]{a_{12}+a_{22}+...+a_{n2}}+...+\sqrt[n]{a_{1p}+a_{2p}+...+a_{np}}$

score 0 · Answer 1

Đặt: $A_{i}=a_{i1}+a_{i2}+...+a_{ip}>0$
$(\forall i=1,2,...,n)$
Theo BĐT Cauchy:
$n\sqrt[n]{\frac{a_{1j}}{A_{1}}\frac{a_{2j}}{A_{2}}...\frac{a_{nj}}{A_{n}}} \leq \frac{a_{1j}}{A_{1}}+\frac{a_{2j}}{A_{2}}+...+\frac{a_{nj}}{A_{n}}$
$(\forall j=1,2,...,p)$
$\Rightarrow n\sum\limits_{j=1}^p\sqrt[n]{\frac{a_{1j}a_{2j}...a_{nj}}{A_{1}A_{2}...A_{n}}} \leq \frac{\sum\limits_{j=1}^pa_{1j}}{A_{1}}+\frac{\sum\limits_{j=1}^pa_{2j}}{A_{2}}+...+\frac{\sum\limits_{j=1}^pa_{nj}}{A_{n}} = n$
$\Rightarrow \sum\limits_{j=1}^p\sqrt[n]{\frac{a_{1j}a_{2j}...a_{nj}}{A_{1}A_{2}...A_{n}}} \leq 1$
$\Rightarrow \sqrt[n]{a_{11}+a_{21}+...+a_{n1}}+\sqrt[n]{a_{12}+a_{22}+...+a_{n2}}+...+\sqrt[n]{a_{1p}+a_{2p}+...+a_{np}}$
$ \leq \sqrt[n]{A_{1}A_{2}...A_{n}} $
$\Rightarrow $ (ĐPCM)

Bài 104245

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Bài 104245

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