=2(xy+yz+zx)

=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======以下答案可供参考======供参考答案1: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。供参考答案2:[(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。

这个答案应该是对的

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