If a × b × v × d × e = 1 and a,b v,d,e are all positive real numbers; what is the minimum value of a+b+c+d+e?

Question

If
abcde=1 (where a,b,c,d and e are all positive real numbers)
then
what is the minimum value of a+b+c+d+e?

Answer 1

if a*b=1 and a and b re +ve (a+b) >2(get by squaring a+b). minimum should be 5 for a+b+c+d+e when all are 1

Answer 2

a*b=1 and a and b re +ve (a+b) >2(get by squaring a+b).
minimum should be 5 for a+b+c+d+e when all are 1

Answer 3

let us start with a*b=1
then b=1/a
a+b=a+(1/a)
differentiate f(a)=a+(1/a)
f '(a)=a-(1/a*a)
equate to zero
a=1,-1 since a is positive a=1, b=1.
differentiate second time
f"(a)=2/(a*a*a)
at a=1 it is positive hence minima
generalize a=b=c=d=e=1