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Dao Trong Luan Coordinator

26/11/2017 at 13:51
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Given: \(\left\{{}\begin{matrix}a+b+c=a^2+b^2+c^2=1\\x:y:z=a:b:c\end{matrix}\right.\)

Prove that: \(\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2}=1\)

 




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    Nguyễn Hưng Phát 26/11/2017 at 14:11

    We have:\(x:y:z=a:b:c\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{x+y+z}{a+b+c}\)

    \(\Rightarrow\dfrac{x^2}{a^2}=\dfrac{y^2}{b^2}=\dfrac{z^2}{c^2}=\dfrac{\left(x+y+z\right)^2}{\left(a+b+c\right)^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\)\(\Rightarrow\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2}=\dfrac{\left(a+b+c\right)^2}{a^2+b^2+c^2}=\dfrac{1^2}{1}=1\)

    \(\Rightarrowđpcm\)

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