Có thể giả thiết \(a\ge b\ge c\). Khi đó : \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{b+c}-\dfrac{1}{2}+\dfrac{b}{c+a}-\dfrac{1}{2}+\dfrac{c}{a+b}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b+a-c}{b+c}+\dfrac{b-c+b-a}{c+a}+\dfrac{c-a+c-b}{a+b}\ge0\)
\(\Leftrightarrow\left(a-b\right)\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)+\left(b-c\right)\left(\dfrac{1}{c+a}-\dfrac{1}{a+b}\right)+\left(c-a\right)\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)\ge0\)
BĐT thức sau cùng đúng với giả thiết ban đầu .
Ta có bài toán phụ: (a+b+c)(\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\))>=9
Từ đó suy ra: (a+b+b+c+c+a)(\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\))>=9
=>(2a+2b+2c)(\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\))>=9
=>\(\left(\dfrac{2a}{a+b}+\dfrac{2b}{a+b}+\dfrac{2c}{a+b}\right)+\left(\dfrac{2b}{b+c}+\dfrac{2a}{b+c}+\dfrac{2c}{b+c}\right)\)+\(\left(\dfrac{2a}{c+a}+\dfrac{2b}{c+a}+\dfrac{2c}{c+a}\right)\)>=9
=>\(\dfrac{2.\left(a+b\right)}{a+b}+\dfrac{2c}{a+b}+\dfrac{2\left(b+c\right)}{b+c}+\dfrac{2a}{b+c}+\dfrac{2\left(c+a\right)}{c+a}+\dfrac{2b}{c+a}\)>=9
=>\(\dfrac{2c}{a+b}+\dfrac{2a}{b+c}+\dfrac{2b}{a+c}\)+2+2+2>=9
=>\(\dfrac{2c}{a+b}+\dfrac{2a}{b+c}+\dfrac{2b}{a+c}\)>=3
=>2\(\left(\dfrac{c}{a+b}+\dfrac{a}{b+c}+\dfrac{b}{c+a}\right)\)>=3
=>\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\)>=\(\dfrac{3}{2}\)
=>đpcm
