Question
If f(x) is a monotonically increasing function " x Î R, f "(x) > 0 and f -1(x) exists, then prove that å{f -1(xi)/3} < f -1({x1+x2+x3}/3), i=1,2,3
Joshi sir comment
f(x) is monotonically increasing so f '(x)>0 and f '' (x) >0 implies that increment of function increases rapidly with increase in x
These informations provide the following informations about the nature of inverse of f(x)
1) f -1 (x) will also be an increasing function but its rate of increase decreases with increasing x
2) for x1 < x2 < x3 , å{f -1(xi)/3} < f -1({x1+x2+x3}/3),
The same result for f(x) will be
å{f (xi)/3} > f ({x1+x2+x3}/3),