2.
Áp dụng BĐT: $\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}$ ta có:
$\dfrac{1}{p-a}+\dfrac{1}{p-b}\ge\dfrac{4}{2p-a-b}=\dfrac{4}{c}$
$\dfrac{1}{p-a}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-a-c}=\dfrac{4}{b}$
$\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge\dfrac{4}{2p-b-c}=\dfrac{4}{a}$
Cộng các BĐT trên lại ta được:
$\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}\ge2(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c})$
Dấu bằng xảy ra khi: $a=b=c$