Ta có : (a^2 + b^2)(x^2 + y^2) = (ax + by)^2 

=> a^2x^2 + a^2y^2 +B^2x^2 + b^2y^2 = a^2x^2 + b^2y^2 + 2axby 

=> chuyển vế trái sang phải: a^2x^2 + a^2y^2 + b^2x^2 + b^2y^2 - a^2x^2 - b^2y^2 - 2axby = 0

=> a^2y^2 + b^2x^2 - 2axby = 0

=> (ax - by)^2 = 0

Chỉ khi ax = by thì (ax - by)^2 = 0 => ax = by. 

Ta có : \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)

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

\(\Leftrightarrow\left(ay\right)^2-2.ay.bx+\left(bx\right)^2=0\)

\(\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\)

Vậy ta có điều phải chứng minh.

Lời giải:
\((a^2+b^2)(x^2+y^2)=(ax+by)^2\)

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

\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)

\(\Leftrightarrow (ay)^2-2(ay)(bx)+(bx)^2=0\)

\(\Leftrightarrow (ay-bx)^2=0\Rightarrow ay=bx\) (đpcm)

Ta có: \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)\)

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

\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)

\(\Leftrightarrow\left(ay-xb\right)^2=0\)

\(\Leftrightarrow ay=xb\)

hay \(\dfrac{a}{x}=\dfrac{b}{y}\)

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

\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)

\(\Leftrightarrow ay=bx\)

\(\Leftrightarrow ay-bx=0\)

( Bất đẳng thức Bu - nhi - a - cốp - xki )

\(a-b=-1\Rightarrow b-a=1\)

\(bx+by-ax-ay\\ =b\left(x+y\right)-a\left(x+y\right)\\ =\left(b-a\right)\left(x+y\right)\\ =1.\left(-2\right)\\ =-2\)