Question

Let a, b, c, d > 0 and $x = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$, then for some a,b,c,d the value of

• A. $x = \frac{1}{2}$
• B. $x = \frac{5}{3}$
• C. $x = \frac{3}{2}$
• D. $x = \frac{5}{4}$

Answer 1

Answer: (C)

3/2

Answer 2

The expression for x is given by $x = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$.

For any positive values of a, b, c, d, each term is positive and less than 1. Summing these gives $0 < x < 4$.

By comparing each term with $\frac{variable}{sum\_of\_all\_variables}$, we showed that $x > 1$.

By comparing each term with $\frac{variable}{sum\_of\_two\_variables}$, we showed that $x < 2$.

So, for any positive a, b, c, d, the value of x is strictly between 1 and 2.

We check the given options. Options B, C, and D are in the range (1, 2). Option A is not.

We try to find if any of the values in B, C, D can be achieved.

Consider the case where $a=c$ and $b=d$. The expression simplifies to $x = \frac{2a}{a+2b} + \frac{2b}{2a+b}$.

We set $x = 3/2$ and $a=1$. This led to a quadratic equation for b, $2b^2 - 7b + 2 = 0$. The roots are $b = \frac{7 \pm \sqrt{33}}{4}$, which are positive real numbers.

Thus, for $a=c=1$ and $b=d=\frac{7 + \sqrt{33}}{4}$ (or $b=d=\frac{7 - \sqrt{33}}{4}$), the value of x is $3/2$.

Since we found positive values for a, b, c, d for which $x=3/2$, this is a possible value.