Hi
onebytwo,
Question 1:
I think this is a celebrated inequality, so am certain that there are probably many different proofs of this out there on the net waiting to be looked up (though I've not seen any myself), but I just had a go at it and here's how I approached it, adhereing to only 4unit methods:
Consider the function: y = x/lnx
__________ [continuous] for all x > 1
We know that x dominates lnx as x -> infinity - i.e. lim(x -> infinity) y = infinity
Consider: dy/dx = 1/lnx - 1/ln
2x
let y' = 0 to find the minimum (note that any
single extremum found must be a minimum for any x > 1 because lim(x -> infinity) y = infinity, so there's no need to perform further derivative tests):
0 = 1/lnx - 1/ln
2x
---> lnx = 1
__________ [how convenient!]
---> x = e
Therefore, x = e > 1 gives a singular extremum and so must be a
minimum.
Hence, since x = pi > 1, then:
y(pi) = pi/ln(pi) > e
---> pi > e.ln(pi)
Raise both sides to 'e' to obtain:
epi > pie __________ as required.
Question 2:
Proving that 1+1 < 5 is not easy if you are looking for a mathematically rigorous proof (much harder than the above proof for the e's and pi's thing).
You can obtain a proof through 1) the order axioms of the field of real numbers, with the natural numbers embedded, along with 2) the definitions and properties of the set of natural numbers,
N, as [the smallest] inductive set.
i.e. the approach would be axiomatic and you will probably need to know some elementary set theory in order to execute a real proof of this inequality.
Hope this helps
.