How do you rationalize the denominator of `1/(sqrt(2)+sqrt(3)+sqrt(5))`?

This is a very good question!

If it were simply `1/(sqrta+sqrtb)`, we would use the conjugate,

and we would multiply by `(sqrta-sqrtb)/(sqrta-sqrtb)`.

Let's try something like that and see if it works. (This is what we do with problems of kinds we have not seen before. Try something and see if it works.)

`1/(sqrt(2)+sqrt(3)+sqrt(5)) = 1/(sqrt(2)+(sqrt(3)+sqrt(5)))`

`= 1/([sqrt(2)+(sqrt(3)+sqrt(5))]) ([sqrt(2)-(sqrt(3)+sqrt(5))])/([sqrt(2)-(sqrt(3)+sqrt(5))])`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))]) / ([sqrt(2) + (sqrt(3)+sqrt(5))][sqrt(2)-(sqrt(3)+sqrt(5))])`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(sqrt(3)+sqrt(5))^2 )`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (2 -(3+2sqrt(15)+5) )`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))]) / (-6 - 2sqrt(15))`

Did that help? (Yes, it did. We now have a more familiar looking problem.

`= ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / ([-6 - 2sqrt(15)][-6 + 2sqrt(15)])`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / (36-4(15))`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))] [-6 + 2sqrt(15)]) / -24`

`= ([sqrt(2)-(sqrt(3)+sqrt(5))] [3 - sqrt(15)]) / 12`

Multiply the numerator if you like, to get:

`= ([sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12`

`=( [sqrt(2)-sqrt(3)-sqrt(5)] [3 - sqrt(15)]) / 12`

`=(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+sqrt45+sqrt75) / 12`

`=(3sqrt(2)-3sqrt(3)-3sqrt(5)- sqrt(30)+3sqrt5+5sqrt3) / 12`

`=(3sqrt(2)+2sqrt(3)- sqrt(30)) / 12`

I've decided to add my version to Jim's as a demonstration or perhaps a warning about the variety of forms the outcome could take:
`1/(sqrt(2)+sqrt(3)+sqrt(5))`

Consider a simpler problem:
If we were asked to rationalize the denominator of
`1/(x+sqrt(5))`

we would simply multiply both the numerator and denominator by the conjugate of the denominator

`1/(x+sqrt(5))xx(x-sqrt(5))/(x-sqrt(5))`

`= (x-sqrt(5))/(x^2-5)`

In this case `x=(sqrt(2)+sqrt(3))`
and the result would be

`=(sqrt(2)+sqrt(3)-sqrt(5))/((sqrt(2)+sqrt(3))^2-5`

`=(sqrt(2)+sqrt(3)-sqrt(5))/(2+2sqrt(2)sqrt(3)+3 -5)`

`=(sqrt(2)+sqrt(3)-sqrt(5))/(2sqrt(2*3))`

In order to complete the rationalization of the denominator we would need to multiply both the numerator and denominator by `sqrt(6)=sqrt(2*3)`

`(sqrt(2)+sqrt(3)-sqrt(5))/(2sqrt(2*3)) xx sqrt(6)/sqrt(2*3)`

`= (sqrt(12)+sqrt(18)-sqrt(30))/(2(2*3))`

`=(2sqrt(3)+3sqrt(2)-sqrt(2*3*5))/12`

I avoided posting this because I am bothered by the result seems to imply that the sequence of the denominator terms is reflected in the result. This should not be true but I have not (yet) been able to come up with a final result where the terms are interchangeable.

If we employ the more general denominator of
`sqrt(a) + sqrt(b)+sqrt(c)`

then we arrive at a numerator of

`(a-b-c)sqrt(a)+(b-a-c)sqrt(b)+(c-a-b)sqrt(c)+2sqrt(abc)`

and a more general (rational) denominator of

`a^2 + b^2 + c^2-2(ab+ac+bc)`

The same work as we see for 2, 3, and 5 yields this result.

In the example given, c = a + b, which eliminates the term that would have contained `sqrt(5)`. Furthermore any time c = a + b, the more general solution simplifies to a numerator of

`bsqrt(a)+asqrt(b)-sqrt(ab(a+b))`
and a denominator of 2ab.

In the specific instance here we have
`(3sqrt(2)+2sqrt(3)-sqrt(30))/12`