#Haskell, 36 bytes

This defines the function # which takes three arguments.

(a#b)c=a*b*c/(b+c-a)/(a+c-b)/(a+b-c) 

You have to call it as follows: (3#4)5

A little bit longer but perhaps more golfable:

p=product f v=p v/p((sum v-).(2*)<$>v) 

Haskell, 36 bytes

This defines the function # which takes three arguments.

(a#b)c=a*b*c/(b+c-a)/(a+c-b)/(a+b-c) 

You have to call it as follows: (3#4)5

A little bit longer but perhaps more golfable:

p=product f v=p v/p((sum v-).(2*)<$>v) 

#Haskell, 36 bytes

This defines the function # which takes three arguments.

(a#b)c=a*b*c/(b+c-a)/(a+c-b)/(a+b-c) 

You have to call it as follows: (3#4)5

#Haskell, 36 bytes

This defines the function # which takes three arguments.

(a#b)c=a*b*c/(b+c-a)/(a+c-b)/(a+b-c) 

You have to call it as follows: (3#4)5

A little bit longer but perhaps more golfable:

p=product f v=p v/p((sum v-).(2*)<$>v)