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Prove the Inequality?
(7!)^(1/7) < (8!)^(1/8)
Here's what I did. First raise to the power 56 on both sides to get:
7!^8 < 8!^7
7!^7 < ( 8!^7) / 7!
7!^7 < ( 8!^6) X 8
630(7!^6) < ( 8!^6)
How can I show it more explicitly??
630(7!^6) < ( 8!^6)
Maybe like this?
630 < 8^6
4 Answers
- RichLv 71 decade agoFavorite Answer
the 7th root of 7! is 3.380015159.....
the 8th root of 8! is 3.7643505.....
so it is true
good luck
- 1 decade ago
Go from 7!^8 < 8!^7
8!^7=7!^7*8^7
7!^8 < 7!^7*8^7
7! < 8^7
From here you could calculate 7! and 8^7. Or you could say that each of these terms is the product of 7 positive integers. Since 8^7 is the product of 7 eights and 7! is the product of 7 integers that are all less than 8, 8^7 is greater than 7!.
8>7>6>5>4>3>2>1
So 8>7
8*8>7*6
8*8*8>7*6*5
and so on.
- 1 decade ago
7!^8 < 8!^7
(7!^7)7! < 8!^7
7! < (8!^7)/(7!^7)
7! < 8^7
which is true bc 7! < 7^7 < 8^7
of course you would start with the bottom statement and work backwards
- ted sLv 71 decade ago
after your 1st step try dividing both sides by [7!]^7....then it is a simple argument to make
