giải giùm mình

Question

cho a,b,c,d>0 thì

$\frac{a+b}{b+c+d}+\frac{b+c}{c+d+a}+\frac{c+d}{d+a+b}+\frac{d+a}{a+b+c}\geq \frac{8}{3}$

cụ nầy lấy đâu ra mấy cái bầy nầy vậy? Chắc chế

score 3 · Answer 1

$VT=\sum \frac{a+b}{b+c+d}=\sum(\frac{a^2}{ab+ac+ad}+\frac{b^2}{bc+bd+ba})\geq \frac{4(a+b+c+d)^2}{4(ab+ac+ad+bc+bd+cd)}=\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd} $

Ta có $(a+b+c+d)^2=a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)$

mà $(a^2+b^2+c^2+d^2)(1+1+1+1)\ge(a+b+c+d)^2$

$\Rightarrow 4(a+b+c+d)^2\ge(a+b+c+d)^2+8(ab+ac+ad+bc+bd+cd)\Leftrightarrow (a+b+c+d)^2\ge\frac83(ab+ac+ad+bc+bd+cd)$

Suy ra đpcm

tran85295 · Answer 2 · 1/27/2016 9:47:45 PM

Bình chọn tăng 4 Bình chọn giảm

$VT=\sum \frac{a+b}{b+c+d}=\sum\frac {a^2}{a(b+c+d)}+\sum\frac{b^2}{b(b+c+d)}$

$ \ge \frac{(a+b+c+d)^2}{2(ab+ac+ad+bc+bd+cd)}+\frac{(a+b+c+d)^2}{2(ab+ac+ad+bc+bd+cd)}=\frac{(a+b+c+d)^2}{ab+ac+ad+bc+bd+cd}=\frac{a^2+b^2+c^2+d^2+2(ab+ac+ad+bc+bd+cd)}{ab+ac+ad+bc+bd+cd}$

$\ge \frac 23 +2=\frac 83$

Trả lời 27-01-16 09:47 PM

tran85295
141 5

10M 559K