Question

Đã trả lời bởi chuyên gia

Answer 1

Lời giải

Ta có: (a² + b² + c²)(x² + y² + z²) = (ax + by + cz)²

⇔ a²x² + a²y² + a²z² + b²y² + b²z² + c²x² + c²y² + c²z² = a²x² + b²y² + c²z² + 2axby + 2axcz + 2bycz

⇔ a²y² + a²z² + b²x² + b²z² + c²x²+ c²y² – 2axby – 2axcz – 2bycz = 0

⇔ (a²y² – 2axby + b²x²) + (a²z² – 2axcz + c²x²) + (b²z² – 2bycz + c²y²) = 0

⇔ (ay – by)² + (az – cx)² + (bz – cy)² = 0

Vì (ay – bx)² ≥ 0; (az – cx)² ≥ 0; (bz – cy)² ≥ 0 nên

(ay – by)² + (az – cx)² + (bz – cy)² ≥ 0

Vậy dấu “=” xảy ra khi:

\(\left\{ \begin{array}{l}ay = bx\\az = cx\\bz = cy\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\frac{a}{x} = \frac{b}{y}\\\frac{a}{x} = \frac{c}{z}\\\frac{b}{y} = \frac{c}{z}\end{array} \right.\)

\( \Rightarrow \frac{a}{x} = \frac{b}{y} = \frac{c}{z}\) (xyz ≠ 0). (đpcm)

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Answer 2