Theo giả thiết: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(\Leftrightarrow\) \(\dfrac{ab+bc+ca}{abc}=0\)
\(\Leftrightarrow\) ab+bc+ca = 0 \(\Rightarrow\) \(\left\{{}\begin{matrix}ab=-ac-bc\\bc=-ab-ac\\ac=-bc-ab\end{matrix}\right.\)
Xét \(\dfrac{a^2}{a^2+2bc}\) = \(\dfrac{a^2}{a^2+bc-ab-ac}\) = \(\dfrac{a^2}{a\left(a-c\right)-b\left(a-c\right)}\)
= \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)
CMTT \(\dfrac{b^2}{b^2+2ac}\) = \(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)
\(\dfrac{c^2}{c^2+2ab}\) = \(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow\) p = \(\dfrac{a^2}{\left(a-b\right)\left(a-c\right)}\)+\(\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\)+\(\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}\)
= \(\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\) = 1
