Ta có : \(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}=6\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)=6.\left(a+b+c\right)\)
\(\Leftrightarrow\frac{\left(a+b+c\right)\cdot\left(a+b\right)}{c}+\frac{\left(a+b+c\right)\cdot\left(b+c\right)}{a}+\frac{\left(a+b+c\right)\cdot\left(c+a\right)}{b}=24\) ( Do \(a+b+c=4\) )
\(\Leftrightarrow\frac{\left(a+b\right)^2+c.\left(a+b\right)}{c}+\frac{\left(b+c\right)^2+a.\left(b+c\right)}{a}+\frac{\left(c+a\right)^2+b.\left(c+a\right)}{b}=24\)
\(\Leftrightarrow\left[\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\right]+2\left(a+b+c\right)=24\)
\(\Leftrightarrow\left[\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}\right]+2.4=24\)
\(\Leftrightarrow\frac{\left(a+b\right)^2}{c}+\frac{\left(b+c\right)^2}{a}+\frac{\left(c+a\right)^2}{b}=16\) ( đpcm )
