Solution 1: Set up the function f(x)=ex/xe, over the domain x>0; we wish to know how f(π) compares with 1.
Using the quotient rule, calculate f ′ (x)=(ex/xe)(1-e/x), which is negative for 0 < x < e, zero at x=e, and positive for x > e.
By the first derivative test, f reaches a unique minimum at x=e, in which case f(e)=1. Thus f(x) > 1 for all other values of x; in particular, for x=π. Thus eπ > πe.
Solution 2: Since the exponential function equals its own derivative, the equation for the straight line tangent to the curve y=ex at (0,1) is y = 1+x
By plugging this value of x into the inequality, multiplying both sides
by e, and then raising both sides to the power of e
(the function y=xe, x > 0, is a strictly increasing function), we obtain eπ > πe.
Comment: Both solutions actually give you more than just the stated inequality; namely they give you ea > ae for any positive a ≠ e.