a) Chứng minh: (ac + bd)2 + (ad – bc)2 = (a2 + b2)(c2 + d2) b) Chứng minh bất dẳng thức Bunhiacôpxki: (ac + bd)2 ≤ (a2 + b2)(c2 + d2)

Giải thích các bước giải:

Ta có:

a,

\(\begin{array}{l}
{\left( {ac + bd} \right)^2} + {\left( {ad - bc} \right)^2}\\
 = {a^2}{c^2} + 2acbd + {b^2}{d^2} + {a^2}{d^2} - 2adbc + {b^2}{c^2}\\
 = {a^2}{c^2} + {b^2}{d^2} + {a^2}{d^2} + {b^2}{c^2}\\
 = \left( {{a^2}{c^2} + {a^2}{d^2}} \right) + \left( {{b^2}{d^2} + {b^2}{c^2}} \right)\\
 = {a^2}\left( {{c^2} + {d^2}} \right) + {b^2}\left( {{c^2} + {d^2}} \right)\\
 = \left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)
\end{array}\)

b,

\(\begin{array}{l}
\left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right) - {\left( {ac + bd} \right)^2}\\
 = {a^2}{c^2} + {a^2}{d^2} + {b^2}{c^2} + {b^2}{d^2} - {a^2}{c^2} - 2acbd - {b^2}{d^2}\\
 = {a^2}{d^2} + {b^2}{c^2} - 2acbd\\
 = {\left( {ad} \right)^2} - 2ad.bc + {\left( {bc} \right)^2}\\
 = {\left( {ad - bc} \right)^2} \ge 0\\
 \Rightarrow {\left( {ac + bd} \right)^2} \le \left( {{a^2} + {b^2}} \right)\left( {{c^2} + {d^2}} \right)
\end{array}\)