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  • 若x,y,z屬于R,a,b,c屬于R+,求證:[(b+c)/a]x^2+[(c+a)/b]y^2+[(a+b)/c]z^2>=2(xy+yz+zx)

    若x,y,z屬于R,a,b,c屬于R+,求證:[(b+c)/a]x^2+[(c+a)/b]y^2+[(a+b)/c]z^2>=2(xy+yz+zx)
    數(shù)學(xué)人氣:196 ℃時間:2020-03-28 22:18:09
    優(yōu)質(zhì)解答
    [(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
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