奥数不等式证明【x+[y+z^(1/4)]^(1/3)】^(1/2)≥(xyz)^(1/32)高手请进,xyz为正数,+ [y + z^(1/4)]^(1/3) 】^(1/2) ≥(xyz)^(1/32),题目不好打,挺挤得的,请见谅.希望能快点.快且对的加分,加满也可以.xyz是(正)
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奥数不等式证明【x+[y+z^(1/4)]^(1/3)】^(1/2)≥(xyz)^(1/32)高手请进,xyz为正数,+ [y + z^(1/4)]^(1/3) 】^(1/2) ≥(xyz)^(1/32),题目不好打,挺挤得的,请见谅.希望能快点.快且对的加分,加满也可以.xyz是(正)
奥数不等式证明【x+[y+z^(1/4)]^(1/3)】^(1/2)≥(xyz)^(1/32)高手请进,
xyz为正数,+ [y + z^(1/4)]^(1/3) 】^(1/2) ≥(xyz)^(1/32),题目不好打,挺挤得的,请见谅.希望能快点.快且对的加分,加满也可以.
xyz是(正)整数还是正数不太确定,如果做不出就默认是正整数吧。望指教,
奥数不等式证明【x+[y+z^(1/4)]^(1/3)】^(1/2)≥(xyz)^(1/32)高手请进,xyz为正数,+ [y + z^(1/4)]^(1/3) 】^(1/2) ≥(xyz)^(1/32),题目不好打,挺挤得的,请见谅.希望能快点.快且对的加分,加满也可以.xyz是(正)
a+b^(1/n) = a+n·[b^(1/n)]/n ≥(n+1)(ab/n^n)^(1/(n+1)),a,b为正数,n为正整数
n^n n+1>(n^n)^(1/(n+1)) => (n+1)/(n^n)^(1/(n+1)) > 1
所以a+b^(1/n) > (ab)^(1/(n+1))
故[x+[y+z^(1/4)]^(1/3)]^(1/2)
>[x+[(yz)^(1/5)]^(1/3)]^(1/2)
=[x+(yz)^(1/15)]^(1/2)
>[(xyz)^(1/16)]^(1/2)
=(xyz)^(1/32)
即[x+[y+z^(1/4)]^(1/3)]^(1/2) > (xyz)^(1/32)
证毕
我知