Có \(\dfrac{a}{b+c}>\dfrac{a}{a+b+c}\)
\(\dfrac{b}{a+c}>\dfrac{b}{a+b+c}\)
\(\dfrac{c}{a+b}>\dfrac{c}{a+b+c}\)
\(\Rightarrow M=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\left(1\right)\)
Lại có a, b, c là 3 cạnh của tam giác
\(\Rightarrow a< b+c;b< a+c;c< a+b\left(BDT\Delta\right)\)
\(\Rightarrow\dfrac{a}{b+c}< 1;\dfrac{b}{a+c}< 1;\dfrac{c}{a+b}< 1\)
\(\dfrac{a}{b+c}< 1\Rightarrow\dfrac{a}{b+c}< \dfrac{a+a}{a+b+c}=\dfrac{2a}{a+b+c}\)
\(\dfrac{b}{a+c}< 1\Rightarrow\dfrac{b}{a+c}< \dfrac{b+b}{a+b+c}=\dfrac{2b}{a+b+c}\)
\(\dfrac{c}{a+b}< 1\Rightarrow\dfrac{c}{a+b}< \dfrac{c+c}{a+b+c}=\dfrac{2c}{a+b+c}\)
\(\Rightarrow M=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}< \dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\)
Từ (1) và (2) suy ra đpcm
