Question

Với a,b,x,y bất kì chứng minh rằng:

(a²⁺b²)*(x²+y²) >= (a*x+b*y)²

Answer 1

(\(a^2\)+\(b^2\)).(\(x^2\)+\(y^2\))>= (ax+by)^2

<=> \(a^2\).\(x^2\)+\(a^2\).\(y^2\)+\(b^2\).\(x^2\)+\(b^2\).\(y^2\)>=\(a^2\).\(x^2\)+2axby+\(b^2\).\(y^2\)

<=> \(a^2\).\(y^2\)- 2aybx+\(b^2\).\(x^2\)>=0

<=> (ay-bx)^2>=0 (luôn đúng)

vậy(\(a^2\)+\(b^2\)).(\(x^2\)+\(y^2\))>=(ax+by)^2

Answer 2

 09/05 lúc 21:40

(a2a2+b2b2).(x2x2+y2y2)>= (ax+by)^2

<=> a2a2.x2x2+a2a2.y2y2+b2b2.x2x2+b2b2.y2y2>=a2a2.x2x2+2axby+b2b2.y2y2

<=> a2a2.y2y2- 2aybx+b2b2.x