áp dung bdt 1/x+1/y>=4/x+y ta co
\(\frac{a+c}{a+b}+\frac{b+d}{b+c}+...\)
=(a+c)(\(\frac{1}{a+b}+\frac{1}{c+d}\)) + (b+d)(\(\frac{1}{b+c}+\frac{1}{a+d}\))\(\ge\)\(\frac{4a+4c}{a+b+c+d}+\frac{4b+4d}{a+b+c+d}\)=4(dpcm)
= \(\left(a+c\right)\left(\frac{1}{a+b}+\frac{1}{c+d}\right)+\left(b+d\right)\left(\frac{1}{b+c}+\frac{1}{d+a}\right)\)
Áp dụng \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\left(x,y>0\right)\)
\(\ge\left(a+c\right)\left(\frac{4}{a+b+c+d}\right)+\left(b+d\right)\left(\frac{4}{a+b+c+d}\right)\)
\(\ge\frac{4\left(a+b+c+d\right)}{a+b+c+d}\)
câu trả lời được Bot lựa Chọn
mà lại được 4 tích sai là sao vậy ?
Bot kém thế à ?
a+c/a+b + b+d/b+c +c+a/c+d + d+d/d+a
=1+c/a+b + 1+ d/b+c + 1+ a/c+d + 1+ b/d+a
=c/a+b + d/b+c + a/c+d +b/d+a +4
vi a,b,c,d>0
nen c/a+b + d/b+c + a/c+d + b/d+a >0
=> c/a+b + d/b+c + a/c+d + b/d+a +4>0
=>a+c/a+b + b+d/b+c + c+a/c+d + d+b/d+a > 4(dpcm)
the nay moi dung bn nhe
ai thay dung thi kik nha
