How to prove xy <= exp(x-1) +ylny using differentiation

[Damned charming :)]

I am teaching a student in a course without partial differentiation so
I can only think of the following method
let b be a constant and set y=b so
xb= exp(x-1) +blnb
means b= exp(x-1) which means x = lnb+1
now d(xb)/dx= b
and d(exp(x-1)+blnb)/dx = exp(x-1)
so if x > lnb+1
xb> exp(x-1) + blnb
since the Right hand side has a higher derivative

so x > lnb+1 Implies xb> exp(x-1) + blnb

so
x > lny +1 implies xy< exp(x-1) + ylny

Now replace x with b
by = exp(b-1) + ylny
means
y=exp(b-1)
now d(by)/dx= b
and d( exp(b-1) + ylny)/dx = 1+ lny
so if lny+ 1 > b
by < exp(b-1) + ylny
since the derivative of the right hand is larger

so lny+1 >x implies xy < exp(x-1)+ylny

 

[fresh_42]

Why not work with the Taylor series of $\exp$ and $\log$?