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已知abc是正实数,求证(b+c)/ax2+(c+a)/by2+(a+b)/cz2≥2(xy+yz+zx)
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已知abc是正实数,求证(b+c)/ax2+(c+a)/by2+(a+b)/cz2≥2(xy+yz+zx)
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答案和解析
(b+c)/ax2+(c+a)/by2+(a+b)/cz2
=(b/a*x2+a/b*y2)+(c/b*y2+b/c*z2)+(a/c*z2+c/a*x2)
≥2xy+2yz+2zx (均值不等式)
=2(xy+yz+zx)
证毕
=(b/a*x2+a/b*y2)+(c/b*y2+b/c*z2)+(a/c*z2+c/a*x2)
≥2xy+2yz+2zx (均值不等式)
=2(xy+yz+zx)
证毕
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