其中2为平方!当p、q都为正数时且p＋q＝1时,试比较（px＋qy）2与px2＋qy2的大小.请问-

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p-1=-q,q-1=-p
(px+qy)²-(px²+qy²)
=p²x²+2pqxy+q²y²-px²-qy²
=px²(p-1)+qy²(q-1)+2pqxy
=pqx²+pqy²+2pqxy
=pq(x²+y²+2xy)
=pq(x+y)²≥0
所以(px+qy)²≥px²+qy²
======以下答案可供参考======
供参考答案1:
(px＋qy)^2－(px^2＋qy^2)
=p(p－1)x^2＋q(q－1)y^2＋2pqxy.
∵p＋q＝1，
∴p－1=－q，q－1=－p.
∴(px＋qy)^2－(px^2＋qy^2)
=－pq(x－y)^2.
∴－pq(x－y)^2≤0.
∴(px＋qy)^2≤px^2＋qy^2