8 > 8^(1/2) + 8^(1/3) + 8^(1/4)
Note that 8 = 2^3 and sqrt(8) < sqrt(9) = 3
8 > 8^(1/2) + 2 + 8^(1/4)
6 = 3 + 3 > sqrt(8) + sqrt(sqrt(8))
Sarthak Chandra <sarthak_loves_math@yahoo.co.in> wrote:
i meant without finding approx values
qcontinuom <lecolin@yahoo.com> wrote:--- In Math4u@yahoogroups.com , Sarthak Chandra
<sarthak_loves_math@...> wrote:
>
> prove
> 5 < 5^(1/2) + 5^(1/3) + 5^(1/4)
> 8 > 8^(1/2) + 8^(1/3) + 8^(1/4)
>
>
>
> --------------------- --------- ---
> Save all your chat conversations. Find them online.
>
Was this all you meant (no variables, no ad infinitum, etc)?
If so, quite straightforward:
As 2.2 < 5^(1/2) and 1.7 < 5^(1/3) and 1.4 < 5^(1/4)
Thus 2.2 + 1.7 + 1.4 is less than the given sum
Yet 5 is less than 5.3 (the sum of these numbers)
Thus by the transitive property QED
Similarly for 8:
As 2.9 > 8^(1/2) and 2.1 > 8 ^(1/3) and 1.7 > 8 ^ (1/4)... etc
But 8 > 6.7, etc
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