a: \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

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

\(=a^2c^2+a^2d^2+b^2d^2+b^2c^2\)

\(=\left(c^2+d^2\right)\left(a^2+b^2\right)\)

b: \(\left(ac+bd\right)^2< =\left(a^2+b^2\right)\left(c^2+d^2\right)\)

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

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

\(\Leftrightarrow\left(ad-bc\right)^2>=0\)(luôn đúng)

a) \(\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

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

\(=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)

\(=\left(a^2c^2+a^2d^2\right)+\left(b^2d^2+b^2c^2\right)\)

\(=a^2\left(c^2+d^2\right)+b^2\left(c^2+d^2\right)\)

\(=\left(a^2+b^2\right)\left(c^2+d^2\right)\)

b) \(\left(a^2+b^2\right)\left(c^2+d^2\right)-\left(ac+bd\right)^{^2}\)

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

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

\(=\left(ad\right)^2-2ad.bc+\left(bc\right)^2\)

\(=\left(ad-bc\right)^2\ge0\)

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

dduungg

hok

tốt 

nha

a) Ta có ${\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$

$=\left(a^2{c}^2+b^2{c}^2\right)+\left(a^2{d}^2+b^2{d}^2\right)$$=c^2\left(a^2+b^2\right)+d^2\left(a^2+b^2\right)$$=\left(a^2+b^2\right)\left(c^2+d^2\right)$

b) Ta có $0\le {\left(ad-bc\right)}^2$$\iff {\left(ac+bd\right)}^2\le {\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2$

Mà theo câu a, ta có ${\left(ac+bd\right)}^2+{\left(ad-bc\right)}^2=\left(a^2+b^2\right)\left(c^2+d^2\right)$

Nên ${\left(ac+bd\right)}^2\le \left(a^2+b^2\right)\left(c^2+d^2\right)$

lom

dasdfghjkl

45ubyu

Có: \(-\left(a-b\right)^2\le0\) với mọi x

=> \(-a^2+2ab-b^2\le0\)

=>\(a^2+2ab+b^2\le2a^2+2b^2\) (cộng cả 2 vế với \(2a^2;2b^2\))

=>\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

\(\Leftrightarrow\left(a+b\right)^2-2\left(a^2+b^2\right)\le0\)

\(\Leftrightarrow-\left(a^2-2ab+b^2\right)\le0\)

\(\Leftrightarrow-\left(a-b\right)^2\le0\)

dấu "=" xẩy ra khi  và chỉ khi a=b

ta có : \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)

<=>\(a^2+2ab+b^2\le2a^2+2b^2\)

<=> \(a^2-2ab+b^2\ge0\)

<=> \(\left(a-b\right)^2\ge0\) bất đẳng thức luôn đúng 

=> ĐPCM

Xét hiệu:

(ac + bd)² - (a² + b²)(c² + d²) = a²c² + 2acbd + b²d² - (a²c² + a²d² + b²c² + b²d²) = - a²d² + 2abcd - b²c²

= - [(ad)² - 2ad.bc + (bc)²] = - (ad - bc)² \(\le\) 0 với mọi a; b; c;d

=> bất đẳng thức cần chứng minh

Dấu "=" xảy ra <=> ad = bc 

Đây là BĐT Bu-nhi-a-cốp-xki mà.