Áp dụng bđt Cauchy-Schwarz:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{a+b}\)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1\right)^2}{b+c}=\dfrac{4}{b+c}\)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{\left(1+1\right)^2}{c+a}=\dfrac{4}{c+a}\)
Cộng theo vế và rút gọn suy ra đpcm
\("="\Leftrightarrow a=b=c\)