Ta có:

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(=a^2x^2-2abxy+b^2y^2+a^2y^2+2abxy+b^2x^2\) \(=\left(ax-by\right)^2+\left(ay+bx\right)\)

\(=vp\)

\(\Rightarrowđpcm\)

Phương Ann Nhã Doanh đề bài khó wá Mashiro Shiina Đinh Đức Hùng

Nguyễn Huy Tú Lightning Farron Akai Haruma

\(\left(a^2+b^2\right)\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(1)

\(\left(ax+by\right)^2+\left(ay-bx\right)^2\)

\(=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)(2)

Từ (1) và (2) ta có \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay-bx\right)^2\)( đpcm )

\(\left(a^2+b^2\right)+\left(x^2+y^2\right)=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

\(\left(ax+by\right)^2+\left(ay-bx\right)^2=a^2x^2+2axby+b^2y^2+a^2y^2-2aybx+b^2x^2\)

\(=a^2x^2+a^2y^2+b^2x^2+b^2y^2\)

Suy ra : \(\left(a^2+b^2\right)+\left(x^2+y^2\right)=\left(ax+by\right)^2+\left(ay+bx\right)^2\left(đpcm\right)\)

VP=\(A^2X^2+B^2Y^2+C^2Z^2+A^2Y^2+B^2X^2+A^2Z^2+C^2X^2+B^2Z^2+C^2Y^2\)

=\(A^2\left(X^2+Y^2+Z^2\right)+B^2\left(X^2+Y^2+Z^2\right)+C^2\left(X^2+Y^2+Z^2\right)\)

=\(\left(X^2+Y^2+Z^2\right)\left(A^2+B^2+C^2\right)\)

a) Sửa đề: \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

Ta có: \(VP=\left(a-b\right)^2+4ab\)

\(=a^2-2ab+b^2+4ab\)

\(=a^2+2ab+b^2\)

\(=\left(a+b\right)^2=VT\)(đpcm)

b) Ta có: \(VT=\left(a-b\right)^2\)

\(=a^2-2ab+b^2\)

\(=a^2+2ab+b^2-4ab\)

\(=\left(a+b\right)^2-4ab=VP\)(đpcm)

c) Ta có: \(VP=\left(ax-by\right)^2+\left(ay+bx\right)^2\)

\(=a^2x^2-2axby+b^2y^2+a^2y^2+2aybx+b^2x^2\)

\(=a^2x^2+b^2y^2+a^2y^2+b^2x^2\)

\(=a^2\left(x^2+y^2\right)+b^2\left(x^2+y^2\right)\)

\(=\left(x^2+y^2\right)\left(a^2+b^2\right)=VT\)(đpcm)

Áp dụng BĐT Bunhiacopxki :

\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

Dấu đẳng thức xảy ra \(\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow ay-bx=0\)

Ta có đpcm.