\(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\)
Áp dụng bất đẳng thức Cauchy - Schwarz dạng phân thức
\(\Rightarrow\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}=\dfrac{9}{2}\)
Vậy \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\ge\dfrac{9}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{3}\)