How to calculate the sum 1/[1+x/(y+z)]+1/[1+y/(z+x)]+1/[1+z/(x+y)]?

We'll calculate the denominator of the 1st
fraction:

1 + x/(y+z) =
(y+z+x)/(y+z)

The inverse of this fraction is
1/(y+z+x)/(y+z) = (y+z)/(x+y+z) (1)

We'll calculate the
denominator of the 2nd fraction:

1 + y/(z+x) =
(z+x+y)/(z+x)

The inverse of this fraction is
1/(z+x+y)/(z+x) = (z+x)/(x+y+z) (2)

We'll calculate the
denominator of the 3rd fraction:

1 + z/(x+y) =
(x+y+z)/(x+y)

The inverse of this fraction is
1/(x+y+z)/(x+y) = (x+y)/(x+y+z) (3)

We'll add (1) + (2) +
(3):

(y+z)/(x+y+z) + (z+x)/(x+y+z) +
(x+y)/(x+y+z)

Since the denominators of the fractions are
matching, we'll re-write the sum such
as:

(y+z+z+x+x+y)/(x+y+z)

We'll
combine like terms inside brackets:

(2x+2y+2z)/(x+y+z) =
2(x+y+z)/(x+y+z)

We'll simplify and we'll
get:

2(x+y+z)/(x+y+z) =
2

The result of the sum of the given
fractions is: 1/[1 + x/(y+z)] + 1/[1 + y/(z+x)] + 1/[1 + z/(x+y)] =
2.