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Ta có:
$\dfrac{(ax+by+cz)^2}{x^2+y^2+z^2}=a^2+b^2+c^2$
$\rightarrow \dfrac{a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz}
{x^2+y^2+z^2}=a^2+b^2+c^2$
Nhân chéo
$a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz=(x^2+y^2+z^2)(a^2+b^2+z^2)$
$\leftrightarrow
a^2x^2+b^2y^2+c^2z^2+2abxy+2acxz+2bcyz=x^2a^2+x^2b^2+x^2c^2+y^2a^2+
y^2b^2+y^2c^2+z^2a^2+z^2b^a+z^2c^2$
$\leftrightarrow
2abxy+2acxz+2bcyz=a^2y^2+a^2z^2+b^2x^2+b^2c^2+c^2x^2+c^2y^2$
$\leftrightarrow 2abxy+2acxz+2bcyz-a^2y^2-a^2z^2-b^2x^2-b^2c^2-c^2x^2-
c^2y^2=0$
$\leftrightarrow (abxy-a^2y^2)-(b^2x^2-abxy)+(acxz-a^2z^2)-(c^2x^2-acxz)+(bcyz-
b^2z^2)-(c^2y^2-bcyz)=0$
$\leftrightarrow ay(bx-ay)-bx(bx-ay)+az(cx-az)-cx(cx-az)+bz(cy-bz)-cy(cy-bz)=0$
$\leftrightarrow (ay-bx)(bx-ay)+(az-cx)(cx-az)+(bz-cy)(cy-bz)=0$
$\leftrightarrow -[(ay-bx)^2+(az-cx)^2+(bz-cy)^2]=0$
$\leftrightarrow (ay-bx)^2+(az-cx)^2+(bz-cy)^2=0$
$(ay-bx)^2 \geq 0 \ \ (az-cx)^2 \geq 0 \ \ (bz-cy)^2 \geq 0$
$\leftrightarrow (ay-bx)^2+(az-cx)^2+(bz-cy)^2 \geq 0$
Dấu "=" xảy ra khi
$ay-bx=0 \ \ az-cx=0 \ \ bz-cy=0$
Hay
$ay=bx \ \ az=cx \ \ bz=cy$
$ay=bx \leftrightarrow \dfrac{a}{x}=\dfrac{b}{y}(1)$
$az=cx \leftrightarrow \dfrac{a}{x}=\dfrac{c}{z}(2)$
Từ (1) và (2) ta được $\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}(dpcm)$
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