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求证:π(X)+π(Y)≥π(X+Y),(X>1,Y>1),注意!因为单位(素数)包含在偶数中,因此该条件多余)
证
由中华单位个数定理知:
Mn+12(√Mn-1)
(1) π(Mn)=-------------, 令Mn≤10^3,Am≤8
Am Mn≥10^3,Am=√Mn-1
1)当Mn<10^3时
X+12(√X-1) X+12√X-12
(2)π(X)=-----------=-------------
8 8
同理
Y+12√Y-12
(3)π(Y)=--------------
8
X+Y+12(√X+√Y)-24
(4)π(X)+π(Y)=--------------------
8
X+Y+12(√X+Y-1) X+Y+12(X+Y)^1/2-12
(5)π(X+Y)=-------------------=-------------------
8 8
若原式成立:
则
X+Y+12(√X+√Y)-24 X+Y+12√X+Y-12
--------------------≥-----------------,两边乘以8得
8 8
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X+Y+12(√X+√Y)-24≥X+Y+12(√X+Y)-12,合并同类项得:
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(6)√X+√Y≥√X+Y+1
当X≥4,Y≥4上式成立
因为 √4+√4=4>[√X+Y+1]=3
√100+√100=20>[√200+1]=15
√400+√400=40>[√800+1]=29
2)当Mn≥10^3,Am=√Mn-1,
X+12(√X-1)
(7)π(X)=-------------=√X+13
√X-1
Y+12(√Y-1)
(8)π(Y)=-------------=√Y+13
√Y-1
X+Y+12(√X+Y-1) ---
(9)π(X+Y)=----------------=√X+Y+13
√X+Y-1
由题意知:
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√X+√Y+26≥√X+Y+13
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(10)√X+√Y≥√X+Y-13
--- --- ---
因为√X+√Y≥√X+Y+1,√X+Y+1≥√X+Y-13
---
所以(10))√X+√Y≥√X+Y-13.
定理证毕.
敬请批评指教!
谢谢!
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