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

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{y}{b}=\dfrac{z}{c}\\\dfrac{x}{a}=\dfrac{z}{c}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ay=bx\\bz=cy\\az=cx\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}ay-bx=0\\bz-cy=0\\az-cx=0\end{matrix}\right.\)

\(\Leftrightarrow\left(ax-by\right)^2+\left(bz-cy\right)^2+\left(az-cx\right)^2=0\)

\(\Leftrightarrow\left(a^2x^2-2axby+b^2y^2\right)+\left(b^2z^2-2bzcy+c^2y^2\right)+\left(a^2z^2-2azcx+c^2x^2\right)=0\)

\(\Leftrightarrow a^2x^2+b^2x^2+c^2x^2+a^2y^2+b^2y^2+c^2y^2+a^2z^2+b^2z^2+c^2z^2-\left(a^2x^2+b^2b^2+c^2y^2+2axby+2azcx+2bzcy\right)=0\)

\(\Leftrightarrow x^2\left(a^2+b^2+c^2\right)+y^2\left(a^2+b^2+c^2\right)+z^2\left(a^2+b^2+c^2\right)-\left(ax+ab+cz\right)^2=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)-\left(ax+by+cz\right)^2=0\)

\(\Leftrightarrow\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

Ta có : \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\) ( theo bđt Bu-nhi-a Cop-xki )

Dấu "=" xảy ra khi \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

Vậy nếu \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) thì \(\left(x^2+y^2+z^2\right)\left(a^2+b^2+c^2\right)=\left(ax+by+cz\right)^2\)

Áp dụng Bunyakovsky:

\(\left(ax+by+cz\right)^2\le\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

Dấu "=" khi: \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) hay \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\) thì thỏa mãn đẳng thức

p/s: Tham khảo,vì t biết lớp 8 chưa học Bunyakovsky,đúng ko Phùng Khánh Linh

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)

Ta co

\(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acz^2-2aczx+acx^2+abx^2-2abxy+aby^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2\left(axby+bcyz+axcz\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(c^2z^2+acz^2+bcz^2\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)}\)

\(=\dfrac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\dfrac{1}{a+b+c}\) ( dpcm)

Phương Ann Nhã Doanh Đinh Đức Hùng Mashiro Shiina

Nguyễn Thanh Hằng Nguyễn Huy Tú Lightning Farron

Akai Haruma Võ Đông Anh Tuấn

mấy anh chị cm cho e thêm cái : \(\dfrac{ay+bx}{c}=\dfrac{bz+cy}{a}=\dfrac{cx+az}{b}\)