Ta có:(a+b+c)2=a2+b2c2+2ab+2bc+2ac
=>ab+bc+ac=0=>ab+ac+bc/abc=0
=>1/a+1/b+1/c=0
=>1/a3+1/b3+1/c3=3/abc
=>bc/a2+ac/a2+ab/c2=abc(1/a3+1/b3+1/c3)=3
theo giả thiết $\left(a+b+c\right)^2=a^2+b^2+c^2$ suy ra ab+ac+bc=0
do đó \(\frac{ab+ac+bc}{abc}=0\) hay \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) suy ra \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\frac{1}{a^3}+\frac{1}{b^3}=-\frac{1}{c^3}\)
\(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\frac{1}{a^3}+\frac{1}{b^3}+3\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(=-3\frac{1}{ab}.\left(\frac{-1}{c}\right)\)
\(=\frac{3}{abc}\)
$\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}$
\(=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
