Ta có: \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{a}{2b}+\dfrac{b}{2c}+\dfrac{c}{2a}=\dfrac{1}{2}\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\ge\dfrac{1}{2}.3=\dfrac{3}{2}\) ( BĐT AM - GM )
Dấu " = " khi a = b = c
\(\Rightarrowđpcm\)
BĐT\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{\left(b+c\right)\left(c+a\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{\left(c-a\right)\left(c+a\right)+\left(a-c\right)\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(b+c\right)\left(c+a\right)}\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)
