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

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### 2 Answers

We have to find the sum 1/[1+x/(y+z)]+1/[1+y/(z+x)]+1/[1+z/(x+y)]

1/[1+x/(y+z)]+1/[1+y/(z+x)]+1/[1+z/(x+y)]

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

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

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

=> 2

**The required sum of 1/[1+x/(y+z)]+1/[1+y/(z+x)]+1/[1+z/(x+y)] is 2.**

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.**