已知a,b,c,x,y,z都是正数,求(b c)/ax^2 (c a)/by^2 (a b)/cz^2>=2(xy yz

题目是不是这样：(b+c)/ax^2+(c+a)/by^2+(a+b)/cz^2≥2(xy+yz+zx)
(b+c)/ax^2+(c+a)/by^2+(a+b)/cz^2
=(b/a*x^2+a/b*y^2)+(c/b*y^2+b/c*z^2)+(a/c*z^2+c/a*x^2)
≥2xy+2yz+2zx (均值不等式)
=2(xy+yz+zx)