若x,y,z属于R,a,b,c属于R+,求证：[(b+c)/a]x^2+[(c+a)/b]y^2+[(a+b)/c]z^2>=2(xy+yz+zx)
人气：397 ℃ 时间：2020-03-28 22:18:09
解答
[(b+c)/a]x^2+[(c+a)/b]y^2+[(a+b)/c]z^2
=(b/a)x^2+(a/b)y^2+(c/a)x^2+(a/c)z^2+(c/b)y^2+(b/c)z^2
=[(b/a)x^2+(a/b)y^2-2xy]+[(c/a)x^2+(a/c)z^2-2xz]+[(c/b)y^2+(b/c)z^2-2yz]
=[(b/a)x^2+(a/b)y^2-2(sqrt(b)x/sqrt(a))(sqrt(a)y/sqrt(b))]
+[(c/a)x^2+(a/c)z^2-2(sqrt(c)x/sqrt(a))(sqrt(a)z/sqrt(c)]
+[(c/b)y^2+(b/c)z^2-2(sqrt(c)y/sqrt(b))(sqrt(b)z/sqrt(c))]
=[(sqrt(b)x/sqrt(a)-(sqrt(a)y/sqrt(b)]^2
+[(sqrt(c)x/sqrt(a)-(sqrt(c)z/sqrt(a)]^2
+[(sqrt(c)y/sqrt(a)-(sqrt(a)z/sqrt(c)]^2
≥0
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