=2(xy+yz+zx)

网友回答

[(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======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。