hihiiii! Chụp đc bài ni !Típ nè!
Bài 18
Suppose $a,b,c\in \mathbb R^+$. Prove that :$(\dfrac ab+\dfrac bc+\dfrac ca)^2\geq (a+b+c)(\dfrac1a+\dfrac1b+\dfrac1c)$
p\s: Bài này không khó lém!
$(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a})^2$ $(a + b + c)(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c})$
$\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} + \dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b}$ $3 + \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$
Dễ dàng chứng minh đc
$\dfrac{a}{c} + \dfrac{b}{a} + \dfrac{c}{b}$ $3$
$ (\dfrac{a^2}{b^2} + 1) + (\dfrac{b^2}{c^2} + 1) + (\dfrac{c^2}{a^2} + 1)$ $2(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a})$
mà $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ $3$
$\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} + \dfrac{c^2}{a^2}$ $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$
Bài viết đã được chỉnh sửa nội dung bởi perfectstrong: 09-08-2011 - 22:26