Verified answer

(e-π)^2 > 0

=> e^2+2eπ+π^2 - 4eπ > 0

=> (e+π)^2 > 4eπ

Take square root,

(e+π) > 2√(eπ)

How to do this without a calculator:

Fact: for two different numbers x and y greater than 1, √(xy) is always smaller than (x+y)/2. So √(eπ) must be smaller than (e+π)/2, which means (e+π) must be greater than 2√(eπ).

Assume 2√(eπ) > (e+π).

Squaring is valid since these are both clearly +ve:

0 > e²+ π² - 2eπ

0 > (e-π)²

Which is clearly bollocks as the square of any real is positive. Note this works for any two distinct real numbers, not just e and π

squaring both, we get

4eπ and

e^2 + π^2 + 2eπ

now we need to compare

(e^2 + π^2) to (2eπ)

but we know both

(π-e)^2 > 0

and

(e-π)^2 > 0

e^2 + π^2 - 2eπ > 0

we get

e^2 + π^2 > 2eπ

working back,

e^2 + π^2 + 2eπ > 4eπ

(e+π)^2 > (2√(eπ))^2

answer :

e+π > 2√(eπ)

(e+π) by just a bit:

(e+π) = 5.8599

2√(eπ) = 5.8446

Nice one!!

e+pie is larger by 0.0153097

Please Mark This Answer As the Best If It was Helpful

Get a calculator.

am i really supposed to know that answer?