Có: [tex](a+b+c)^2=a^2+b^2+c^2 \\\rightarrow a^2+b^2+c^2+2ab+2bc+2ca=a^2+b^2+c^2\\\rightarrow 2ab+2bc+2ca=0\\\rightarrow ab+bc+ca=0\\\rightarrow \frac{ab+bc+ca}{abc}=0 \\\rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0[/tex]
Mặt khác
[tex]\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{cba}{b^3}+\frac{abc}{c^3}=abc.(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3})[/tex]
Giờ dùng HĐT [tex]a^3 + b^3 + c^3 = ((a + b + c)(a² + b²+ c²-ab-bc-ca)+3abc)[/tex] có:
[tex]abc.(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3})=abc((\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca})+\frac{3}{abc})=abc(0+\frac{3}{abc})=abc.\frac{3}{abc}=3(dpcm)[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
