nhầm làm lại nha ^^

(a+b+c)^2=a^2+b^2+c^2

=>a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

=>2(ab+bc+ac)=0

=>ab+bc+ac=0

=>(ab+bc+ac)/abc=0

=>ab/abc+bc/abc+ac/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 (đpcm)

(a+b+c)^2=a^2+b^2+c^2

a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2

2(ab+bc+ac)=0

ab+bc+ac=0

(ab+bc+ac)/abc=0

ab/abc+bc/abc+ac/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.(1/a.)(1/b).(1/a+1/b)=-1/c^3

=> 1/a^3+1/b^3+1/c^3.3ab.(-1/c)=0

=> 1/a^3+1/b^3+1/c^3=3/abc

Ta có: \(a^2+bc\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)\(\Rightarrow\frac{1}{a^2+bc}\le\frac{1}{2a\sqrt{bc}}\)

Tương tự ta có:

\(\frac{1}{b^2+ac}\le\frac{1}{2b\sqrt{ac}};\frac{1}{c^2+ab}\le\frac{1}{2c\sqrt{ab}}\)

Cộng theo vế ta có:

\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{1}{2a\sqrt{bc}}+\frac{1}{2b\sqrt{ac}}+\frac{1}{2c\sqrt{ab}}\)

\(\Leftrightarrow\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{\sqrt{bc}}{2abc}+\frac{\sqrt{ac}}{2abc}+\frac{\sqrt{ab}}{2abc}\)

\(\Leftrightarrow\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\le\frac{\sqrt{bc}+\sqrt{ac}+\sqrt{ab}}{2abc}\le\frac{a+b+c}{2abc}\)

Đẳng thức xảy ra khi \(a=b=c\)

ai cứu mình với ạ:(

2.3+3.(-1,2)+(-1,2).2=0 (a=2, b=3, c=-1,2)

\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=\dfrac{19}{18}\)

\(\dfrac{3}{abc}=-\dfrac{5}{12}\)?