Ta có :\(\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}>\frac{a}{a+b+c+d}+\frac{b}{a+b+c+d}+\frac{c}{a+b+c+d}+\frac{d}{a+b+c+d}=\frac{a+b+c+d}{a+b+c+d}=1\)và \(\frac{a}{b+c+d}+\frac{b}{c+d+a}+\frac{c}{d+a+b}+\frac{d}{a+b+c}<\frac{2a}{a+b+c+d}+\frac{2b}{a+b+c+d}+\frac{2c}{a+b+c+d}+\frac{2d}{a+b+c+d}=\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)

Đặt A=a/b+c+d + b/c+d+a +c/d+a+b +d/a+b+c

     4+A=a/b+c+d +1  + b/c+d+a  +1 + c/d+a+b  +1  + d/a+b+c  +1

     4+A=2a/a+b+c+d  +  2b/a+b+c+d  +  2c/a+b+c+d  +2d/a+b+c+d

     4+A=2a+2b+2c+2d/a+b+c+d

     4+A=2(a+b+c+d) /a+b+c+d

     4+A=2

       A=2-4= -2

=) A<1<2

m = 3/2 = 1.5 >1

kich mk di

diem mk thap qua

thank you

Ta có: m=\(\frac{a}{c+b}+\frac{b}{c+a}+\frac{c}{a+b}\)

              = 1/2 <1

a) Ta có: a<b

\(\Leftrightarrow ac< bc\)

\(\Leftrightarrow ac+ab< bc+ab\)

\(\Leftrightarrow a\left(b+c\right)< b\left(a+c\right)\)

hay \(\frac{a}{b}< \frac{a+c}{b+c}\)(đpcm)

b) Ta có: \(\frac{a}{a+b}>\frac{a}{a+b+c}\)

\(\frac{b}{b+c}>\frac{b}{a+b+c}\)

\(\frac{c}{c+a}>\frac{c}{a+b+c}\)

Cộng vế theo vế, ta được:

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a+b+c}{a+b+c}=1\)

hay \(1< \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)(1)

Ta có: \(\frac{a}{a+b}< \frac{a+c}{a+b+c}\)

\(\frac{b}{b+c}< \frac{b+a}{a+b+c}\)

\(\frac{c}{c+a}< \frac{c+b}{a+b+c}\)

Cộng vế theo vế, ta được:

\(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c+b+c+a+b}{a+b+c}=2\)(2)

Từ (1) và (2) suy ra \(1< A< 2\)

hay A không phải là số nguyên(đpcm)

sai sai

Đáp số là M > 1. Bạn cần cách giải không ?

Co minh biet ket qua roi ban HiHI

mk chi cac bạn tuyet chieu;

nhung bai toan dang nay mk thuong lay so cu the nhu 1;2;3 .... thay vao se doan dc kq vi violympic thoi gian thi co hạn

cac ban co dong y k

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