If $\frac{x}{y+z} = a; \frac{y}{z+x} = b$ and $\frac{z}{x+y} = c$, then $\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}$ is equal to

### Concept & Formula This problem requires recognizing symmetry in algebraic fractions. The key insight is to transform the isolated variables $a$, $b$, and $c$ into expressions that share a common denominator when manipulated. Adding $1$ to each fractional equation creates a uniform numerator of $(x + y + z)$. $$ \text{Symmetric Manipulation: } \frac{u}{v+w} + 1 = \frac{u+v+w}{v+w} $$ ### Step-by-Step Solution * **Given:** * $\frac{x}{y+z} = a$ * $\frac{y}{z+x} = b$ * $\frac{z}{x+y} = c$ * Target: $\frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c}$ * **Calculation:** 1. Let us manipulate the first given equation to find the value of $1+a$. 2. Add $1$ to both sides: $1 + a = 1 + \frac{x}{y+z}$. 3. Take the common denominator: $1 + a = \frac{y+z+x}{y+z}$. 4. The target expression requires $\frac{1}{1+a}$, which is the reciprocal. 5. Reciprocal: $\frac{1}{1+a} = \frac{y+z}{x+y+z}$. 6. Apply the exact same logic symmetrically to $b$ and $c$: * $\frac{1}{1+b} = \frac{z+x}{x+y+z}$ * $\frac{1}{1+c} = \frac{x+y}{x+y+z}$ 7. Substitute these reciprocal values into the target expression: $\frac{y+z}{x+y+z} + \frac{z+x}{x+y+z} + \frac{x+y}{x+y+z}$ 8. Since the denominators are identical, simply add the numerators: $\frac{y+z + z+x + x+y}{x+y+z}$ 9. Group like terms in the numerator: $\frac{2x + 2y + 2z}{x+y+z}$. 10. Factor out the $2$: $\frac{2(x+y+z)}{x+y+z}$. 11. Cancel the common term $(x+y+z)$ to get $2$. ### Exam Strategy & Shortcut For highly symmetric algebraic problems where variables are cyclical and no absolute values are given, you can use the Value Assumption method. Assume $x = y = z = 1$. Then, $a = \frac{1}{1+1} = \frac{1}{2}$. By symmetry, $b = \frac{1}{2}$ and $c = \frac{1}{2}$. Now evaluate the target: $\frac{1}{1+0.5} + \frac{1}{1+0.5} + \frac{1}{1+0.5} = \frac{1}{1.5} + \frac{1}{1.5} + \frac{1}{1.5} = \frac{2}{3} + \frac{2}{3} + \frac{2}{3} = \frac{6}{3} = 2$. This completely bypasses all algebraic manipulation! ### Common Pitfall Students often attempt to cross-multiply the original equations to isolate $x$, $y$, and $z$, and then substitute them into the target expression. This leads to a massive, tangled web of equations that is nearly impossible to solve within the time limits of a competitive exam. Always look for structural symmetry first. ### Final Answer Therefore, the correct answer is 2.