bài này khó quá hay bạn gửi lên ban quản lý Onine Math
\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=a^2+b^2+c^2\)
\(\Leftrightarrow2\left(ab+bc+ca\right)=0\Leftrightarrow ab+bc+ca=0\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{0}{abc}\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{1}{b}+\frac{1}{c}=\frac{-1}{a}\Leftrightarrow\left(\frac{1}{b}+\frac{1}{c}\right)^3=\frac{-1}{a^3}\)
\(\Leftrightarrow\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{bc}\left(\frac{1}{b}+\frac{1}{c}\right)=\frac{-1}{a^3}\Leftrightarrow\frac{1}{b^3}+\frac{1}{c^3}+\frac{3}{bc}.\frac{-1}{a}=-\frac{1}{a^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\Leftrightarrow abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=3\Leftrightarrow\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=3\)
Ta có đpcm.
