Question

[Hàn Kim Băng]

a) Chứng minh: (ac + bd)² + (ad – bc)² = (a² + b²)(c² + d²)

b) Chứng minh bất đẳng thức Bunhiacôpxki: (ac + bd)² ≤ (a² + b²)(c² + d²)

Answer 1

[kudo shinichi]

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2+2abcd-2abcd=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(\Leftrightarrow a^2c^2+b^2d^2+a^2d^2+b^2c^2-a^2c^2-a^2d^2-b^2c^2-b^2d^2=0\)

\(\Leftrightarrow0=0\)( luôn đúng )

​​Vậy \(\left(ac+bd\right)^2+\left(ad-bc\right)^2=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

b) \(\left(ac+bd\right)^2\le\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(\Leftrightarrow a^2c^2+b^2d^2+2abcd\le a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

\(\Leftrightarrow a^2d^2+b^2c^2-2abcd\ge0\)

\(\Leftrightarrow\left(ad-bc\right)^2\ge0\)( luôn đúng )

Dấu " = " xảy ra \(\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)