Question

当pq都为正数且p+q=1时,试比较代数式（px+qy）^2与px^2+qy^2的大小

Answer 1

px^2+qy^2-（px+qy）^2=px^2+qy^2-(px)^2-(qy)^2-2pqxy

=(p-p^2)x^2+(q-q^2)y^2-2pqxy

=pqx^2+pqy^2-2pqxy

=pq(x^2-2xy+y^2)

=pq(x-y)^2

故当x等于y时 .

当x不等于y时 .

又PQ为正数.

故 （px+qy）^2小于等于 px^2+qy^2