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abc为正数 a+b=1 求证axx+byy>或等于(ax+by)(ax+by)
人气:296 ℃ 时间:2020-04-16 05:11:34
解答
ax^2+by^2-(ax+by)^2
=ax^2+by^2-(a^2x^2+2abxy+b^2y^2)
=a(1-a)x^2-2abxy+b(1-b)y^2
=abx^2-2abxy+bay^2
=ab(x^2-2xy+y^2)
=ab(x-y)^2
>=0
所以ax^2+by^2>=(ax+by)^2
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