\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{a}{a+b}-\dfrac{1}{2}+\dfrac{b}{b+c}-\dfrac{1}{2}+\dfrac{c}{c+a}-\dfrac{1}{2}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}+\dfrac{b-a+a-c}{2\left(b+c\right)}+\dfrac{c-a}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2\left(a+b\right)}-\dfrac{a-b}{2\left(b+c\right)}+\dfrac{a-c}{2\left(b+c\right)}-\dfrac{a-c}{2\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\left(\dfrac{1}{a+b}-\dfrac{1}{b+c}\right)+\dfrac{a-c}{2}\left(\dfrac{1}{b+c}-\dfrac{1}{c+a}\right)\ge0\)
\(\Leftrightarrow\dfrac{a-b}{2}\cdot\dfrac{c-a}{\left(a+b\right)\left(b+c\right)}+\dfrac{a-c}{2}\cdot\dfrac{a-b}{\left(b+c\right)\left(c+a\right)}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)}{2}\left(\dfrac{1}{\left(b+c\right)\left(c+a\right)}-\dfrac{1}{\left(a+b\right)\left(b+c\right)}\right)\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{2\left(a+b\right)\left(a+c\right)\left(b+c\right)}\ge0\)(luôn đúng)
\(\Rightarrowđpcm\)
