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Question

a) Cmr: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}$
b) Chứng minh dạng tổng quát:
$\frac{1}{a_1}+\frac{1}{a_2}+\frac{1}{a_3}+...+\frac{1}{a_n}\geq\frac{n^2}{a_1+a_2+a_3+...+a_n}$

score 5 · Answer 1

Bình chọn tăng 5 Bình chọn giảm

b)
Áp dụng BĐT Cauchy ta có:
$\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\ge n\sqrt[n]{\frac{1}{a_1a_2\ldots a_n}}$
$a_1+a_2+\ldots+a_n\ge n\sqrt[n]{a_1a_2\ldots a_n}$
Suy ra:
$(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n})(a_1+a_2+\ldots+a_n)\ge n^2\Leftrightarrow \frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_n}\ge\frac{n^2}{a_1+a_2+\ldots+a_n}$
Dấu bằng xảy ra khi: $a_1=a_2=\ldots=a_n$

đơn giản nhỉ – cuungonghinh 25-11-12 09:19 PM

score 4 · Answer 2

Bình chọn tăng 4 Bình chọn giảm

a)
Áp dụng BĐT Cauchy ta có:
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}$
$a+b+c\ge3\sqrt[3]{abc}$
Suy ra: $(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)\ge9\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}$
Dấu bằng xảy ra khi: $a=b=c$