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 x< 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), 

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