摘要:证一:∵a, b, c, d, x, y都是正数 ∴要证:xy≥ac + bd 只需证:(xy)2≥(ac + bd)2 即:(a2 + b2)(c2 + d2)≥a2c2 + b2d2 + 2abcd 展开得:a2c2 + b2d2 + a2d2 + b2c2≥a2c2 + b2d2 + 2abcd 即:a2d2 + b2c2≥2abcd 由基本不等式.显然成立 ∴xy≥ac + bd 证二:xy = ≥ 证三: ∵x2 = a2 + b2.∴不妨设a = xsina, b = xcosa y2 = c2 + d2 c = ysinb, d = ycosb ∴ac + bd = xysinasinb + xycosacosb = xycos≤xy

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