\(M=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)
\(=\left(1-\frac{b}{a+b}\right)+\left(1-\frac{c}{b+c}\right)+\left(1-\frac{a}{c+a}\right)\)
\(< 3-\left(\frac{b}{a+b+c}+\frac{c}{b+c+a}+\frac{a}{c+a+b}\right)=3-1=2\)
=>M < 2
Ta có:
\(\frac{a}{a+b}<\frac{a+c}{a+b+c};\frac{b}{b+c}<\frac{b+a}{a+b+c};\frac{c}{c+a}<\frac{c+b}{a+b+c}\)
=> \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}<\frac{a+c}{a+b+c}+\frac{b+a}{a+b+c}+\frac{c+b}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Vậy M < 2.
M=a/(a+b)+b/(b+c)+c/(c+a)
=(a+b-b)/(a+b)+(b+c-c)/(b+c)+(c+a-a)/(c+a)
=3-[(b/(a+b)+c/(b+c)+a/(c+a)]
Ta có:b/(a+b)>b/(a+b+c)
c/(b+c)>c/(a+b+c)
a/(c+a)>a/(a+b+c)
=>M>3-[(a/(a+b+c)+b/(a+b+c)+c/(a+b+c)]=2
Vậy M>2
Với \(a,b,c\in N\)*
Ta có: Nếu \(\frac{a}{b}< 1\)thì \(\frac{a+c}{b+c}< 1\)
* Áp dụng tính chất này, ta có: - Vì \(\frac{a}{a+b}< 1\)\(\Rightarrow\frac{a}{a+b}< \frac{a+c}{a+b+c}\)
- Vì \(\frac{b}{b+c}< 1\)\(\Rightarrow\frac{b}{b+c}< \frac{b+a}{b+c+a}\)
- Vì \(\frac{c}{c+a}< 1\)\(\Rightarrow\frac{c}{c+a}< \frac{c+b}{c+a+b}\)
* Cộng từng vế các đẳng thức trên, ta có: \(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+c}{a+b+c}+\frac{b+a}{b+c+a}+\frac{c+b}{c+a+b}\)
\(=\frac{a+c+b+a+c+b}{a+b+c}=\frac{2.\left(a+b+c\right)}{a+b+c}=2\)
Vậy \(M< 2\)
