Question

[Hồ Khánh Châu]

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{9}{a+b+c}=0\)

\(\frac{bc}{abc}+\frac{ac}{bca}+\frac{ab}{cab}-\frac{9abc}{\left(a+b+c\right)abc}=0\)

\(\left(A+b+c\right)bc+\left(a+b+c\right)ac+\left(a+b+c\right)ab-9abc=0\)

\(b^2c+c^2b+abc+a^2c+c^2a+abc+a^2b+b^2a+abc-9abc=0\)

\(b^2c+c^2b+a^2c+c^2a+a^2b+b^2a-6abc=0\)

\(c\left(b^2+a^2\right)+b\left(c^2+a^2\right)+a\left(c^2+b^2\right)-6abc=0\)

\(c\left(b^2+a^2-2ab\right)+b\left(c^2-2ac+a^2\right)+a\left(c^2+2cb+b^2\right)=0\)

\(c\left(b-a\right)^2+b\left(c-a\right)^2+a\left(c-b\right)^2=0\)

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