Ta có (a+b+c)^2=a^2+b^2+c^2
-> a^2+b^2+c^2 +2ab+2bc+2ca=a^2+b^2+c^2
-> 2(ab+bc+ca)=0 -> ab+bc+ca=0
-> (ab+bc+ca)/abc=0
-> ab/abc+bc/abc+ca/abc=0
-> 1/c+1/a+1/b=0
-> 1/a+1/b=-1/c
->(1/a+1/b)^3=(-1/c)^3
-> 1/a^3+1/b^3+3/ab.(1/a+1/b)=-1/c^3
->1/a^3+1/b^3+1/c^3+3/ab.(-1/c)=0
-> 1/a^3+1/b^3+1/c^3-3/abc=0
->1/a^3+1/b^3+1/c^3=3/abc
-> (b^3.c^3+a^3.c^3+a^3.b^3)/a^3.b^3.c^3=3/abc( quy đồng lên nhé)
-> (b^3.c^3+a^3.c^3+a^3.b^3)/a^2.b^2.c^2 . 1/abc=3/abc
->(b^3.c^3)/(a^2.b^2.c^2)+(a^3.c^3)/(a^2.b^2.c^2)+(a^3.b^3)/(a^2.b^2.c^2)=3
-> bc/a^2+ac/b^2+ab/c^2=3 (dcpcm)