高中不等式证明((b+c)/a)x^2+((c+a)/b)y2+((a+b)/c)z2≥2(xy+y

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高中不等式证明((b+c)/a)x^2+((c+a)/b)y2+((a+b)/c)z2≥2(xy+yz+zx)x,y,z是实数a,b,c是正实数求证((b+c)/a)x^2+((c+a)/b)y^2+((a+b)/c)z^2≥2(xy+yz+zx)(图1)
======以下答案可供参考======
供参考答案1:
((b+c)/a)x^2+((c+a)/b)y^2+((a+b)/c)z^2
=bx^2/a+cx^2/a+cy^2/b+ay^2/b+az^2/c+bz^2/c
=[(b/a)x^2+(a/b)y^2]+[(c/a)x^2+(a/c)z^2]+[(c/b)y^2+(b/c)z^2]
a>0,b>0所以(b/a)x^2=0，(a/b)y^2>=0所以(b/a)x^2+(a/b)y^2>=2根号[(b/a)x^2*(a/b)y^2]=2xy
同理(c/a)x^2+(a/c)z^2>=2xz(c/b)y^2+(b/c)z^2>=2yz相加即得((b+c)/a)x^2+((c+a)/b)y^2+((a+b)/c)z^2>=2(xy+yz+zx)
供参考答案2:
((b+c)/a)x^2+((c+a)/b)y^2+((a+b)/c)z^2
=bx^2/a+ay^2/b+cx^2/a+az^2/c+cy^/2b+bz^2/c
≥2(xy+yz+zx)