Áp dụng bđt bunhiacopxki \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
dấu "=" xảy ra \(< =>ay=bx< =>\frac{a}{x}=\frac{b}{y}\)
Ta có: (a2+b2).(x2+y2)=(ax+by)2
<=>a2x2+a2y2+b2x2+b2y2=a2x2+b2y2+2abxy
<=>a2y2+b2x2=2abxy
<=>a2y2+b2x2-2abxy=0
<=>(ay-bx)2=0
<=>ay=bx
<=>\(\frac{a}{x}=\frac{b}{y}\)