Bất đẳng thức này >=3/2!!!!!!!!!!!!!
\(\frac{a}{b+c}+1+\frac{b}{a+c}+1+\frac{c}{a+b}+1-3=\left(a+b+c\right)\cdot\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\)
áp dung cosy ta có \(x+y+z\ge3\sqrt[3]{x\cdot y\cdot z}\) \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge3\sqrt[3]{\frac{1}{x\cdot y\cdot z}}\)
\(\Rightarrow\left(x+y+z\right)\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge9\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)
\(\Rightarrow\left(\frac{1}{b+c}+\frac{1}{a+c}+\frac{1}{a+b}\right)\ge\frac{9}{2\cdot\left(a+b+c\right)}\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\ge\frac{9}{2}\)
\(\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge\frac{9}{2}-3=\frac{3}{2}\)
