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)
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![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)](/uploads/image/z/6960228-60-8.jpg?t=%E8%8B%A5x%2Cy%2Cz%E5%B1%9E%E4%BA%8ER%2Ca%2Cb%2Cc%E5%B1%9E%E4%BA%8ER%2B%2C%E6%B1%82%E8%AF%81%EF%BC%9A%5B%28b%2Bc%29%2Fa%5Dx%5E2%2B%5B%28c%2Ba%29%2Fb%5Dy%5E2%2B%5B%28a%2Bb%29%2Fc%5Dz%5E2%3E%3D2%28xy%2Byz%2Bzx%29)
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)
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)
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)
[(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
证:b/ax²+a/by²≥2√(x²y²)=2|xy|≥2xy
同理c/ax²+a/cz²≥2zx,c/by²+b/cz²≥2yz
上三式相加,即为所证。
[(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...
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[(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
证:b/ax2+a/by2≥2√(x2y2)=2|xy|≥2xy
同理c/ax2+a/cz2≥2zx,c/by2+b/cz2≥2yz
上三式相加,即为所证。
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