Chủ đề này có 1 trả lời

[Dam Uoc Mo]

Cho a,b,c,d>0 và abcd=1.CMR:

$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(d+1)}{1+cd+d}+\frac{(c-1)(a+1)}{1+da+a}+\frac{(d-1)(b+1)}{1+ab+b}\geq 0.$

[Yagami Raito]

Cho a,b,c,d>0 và abcd=1.CMR:

$\frac{(a-1)(c+1)}{1+bc+c}+\frac{(b-1)(d+1)}{1+cd+d}+\frac{(c-1)(a+1)}{1+da+a}+\frac{(d-1)(b+1)}{1+ab+b}\geq 0.$

 

Lời Giải: 

Ta có 

$VT+4=(\dfrac{(a-1)(c+1)}{1+bc+c}+1)+(\dfrac{(b-1)(d+1)}{1+cd+d}+1)+(\dfrac{(c-1)(a+1)}{1+da+a}+1)+(\dfrac{(d-1)(b+1)}{1+ab+b}+1)$

 

$=\dfrac{a+bc+ac}{1+bc+c}+\dfrac{b+cd+bd}{1+cd+d}+\dfrac{c+da+ca}{1+da+a}+\dfrac{d+ab+bd}{1+ab+b} $

 

$=\dfrac{(a+bc+ac)(\dfrac{1}{a}+bc+\dfrac{c}{a})}{(1+bc+c)(\dfrac{1}{a}+bc+\dfrac{c}{a})}+\dfrac{(b+cd+bd)(\dfrac{1}{b}+cd+\dfrac{d}{b})}{(1+cd+d)(\dfrac{1}{b}+cd+\dfrac{d}{b})}+\dfrac{(c+da+ca)(\dfrac{1}{c}+da+\dfrac{a}{c})}{(1+da+a)(\dfrac{1}{c}+da+\dfrac{a}{c})}+\dfrac{(d+ab+bd)(\dfrac{1}{d}+ab+\dfrac{b}{d})}{(1+ab+b)(\dfrac{1}{d}+ab+\dfrac{b}{d})} $

 

$\ge\dfrac{(1+bc+c)^2}{(1+bc+c)(\dfrac{1}{a}+bc+\dfrac{c}{a})}+\dfrac{(1+cd+d)^2}{(1+cd+d)(\dfrac{1}{b}+cd+\dfrac{d}{b})}+\dfrac{(1+da+a)^2}{(1+da+a)(\dfrac{1}{c}+da+\dfrac{a}{c})}+\dfrac{(1+ab+b)^2}{(1+ab+b)(\dfrac{1}{d}+ab+\dfrac{b}{d})} $

 

$=\dfrac{ad(1+bc+c)}{(1+cd+d)}+\dfrac{ba(1+cd+d)}{(1+da+a)}+\dfrac{cb(1+da+a)}{(1+ab+b)}+\dfrac{dc(1+ab+b)}{(1+bc+c)}\ge 4 $ (Theo Cô-si và $abcd=1$)

 

Như vậy BĐT đã chứng minh xong. $\blacksquare$

Bài viết đã được chỉnh sửa nội dung bởi Yagami Raito: 15-07-2014 - 12:26