证明xyz(x+y)(y+z)(x+z)≤(x^2+y^2)(y^2+z^2)(x^2+z^2)
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证明xyz(x+y)(y+z)(x+z)≤(x^2+y^2)(y^2+z^2)(x^2+z^2)
证:利用柯西不等式(a^2+b^2)(c^2 + d^2)≥(ac+bd)^2
得(x^2+y^2)(y^2+z^2)≥(xy + yz)^2≥0
(y^2+z^2)(x^2+z^2)≥(yz + zx)^2≥0
(x^2+z^2)(x^2+y^2)≥(zx + xy)^2≥0
因为三式均为正,故可以相乘.相乘,得
[(x^2+y^2)(y^2+z^2)(x^2+z^2)]^2≥[(xy + yz)(yz + zx)(zx + xy)]^2
因(x^2+y^2)(y^2+z^2)(x^2+z^2)≥0,故可同时开方,得
(x^2+y^2)(y^2+z^2)(x^2+z^2)≥(xy + yz)(yz + zx)(zx + xy)
即xyz(x+y)(y+z)(x+z)≤(x^2+y^2)(y^2+z^2)(x^2+z^2),原式得证.
得(x^2+y^2)(y^2+z^2)≥(xy + yz)^2≥0
(y^2+z^2)(x^2+z^2)≥(yz + zx)^2≥0
(x^2+z^2)(x^2+y^2)≥(zx + xy)^2≥0
因为三式均为正,故可以相乘.相乘,得
[(x^2+y^2)(y^2+z^2)(x^2+z^2)]^2≥[(xy + yz)(yz + zx)(zx + xy)]^2
因(x^2+y^2)(y^2+z^2)(x^2+z^2)≥0,故可同时开方,得
(x^2+y^2)(y^2+z^2)(x^2+z^2)≥(xy + yz)(yz + zx)(zx + xy)
即xyz(x+y)(y+z)(x+z)≤(x^2+y^2)(y^2+z^2)(x^2+z^2),原式得证.
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