(1+X^2)(1+y^2)(1+z^2)>=8xyz求证.麻烦大家教教我.响了好久不懂.嘻嘻

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(1+X^2)(1+y^2)(1+z^2)>=8xyz求证.麻烦大家教教我.响了好久不懂.嘻嘻
(1+X^2)(1+y^2)(1+z^2)>=8xyz
求证.
麻烦大家教教我.响了好久不懂.嘻嘻

(1+X^2)(1+y^2)(1+z^2)>=8xyz求证.麻烦大家教教我.响了好久不懂.嘻嘻
.1+x*2≥2x
1+y*2≥2y
I+z*2≥2z(上式都运均值不等式)
所以乘起来就大于8xyz

不等式公式:a^2+b^2≥2ab
因为
1+x^2≥2x
1+y^2≥2y
1+z^2≥2z
所以:
(1+x^2)(1+y^2)(1+z^2)≥8xyz

由于算术均值>=几何均值
(1+x^2)>=根号(1*x^2)=x,
即(1+x^2)>=2x,
同理(1+y^2)>=2y
(1+z^2)>=2z
即得1+X^2)(1+y^2)(1+z^2)>=8xyz
(应当有条件:x,y,z>0)

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