Question

Chứng minh rằng nếu a.b.c=a+b+c và 1/a+1/b+1/c=2 thì 1/a^2+1/b^2+1/c^2=2

Answer 1

Theo gt, ta có: \(a+b+c=abc\)

\(\Leftrightarrow\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{ab}=1\)

Đặt \(\dfrac{1}{a}=x;\dfrac{1}{b}=y;\dfrac{1}{c}=z\)

\(\Rightarrow\left\{{}\begin{matrix}x+y+z=2\\xy+yz+xz=1\end{matrix}\right.\)

Mặt khác, ta có: \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+xz\right)\)

\(\Rightarrow x^2+y^2+z^2=2^2-2\times1=2\)

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

Vậy ta có đpcm.