The ubiquitous AM-GM inequality has many
applications. It almost crops up in unlikely situations and
the solutions using AM-GM are truly elegant . Recall
that for n positive reals `a_(i) I = 1,2 …,`n, the AM-GM inequality tells
`(overset(n) underset(1)suma_i)/n ge ( overset(n)underset(1)proda_i)^((1)/(n))`
The special in which the inequality turns into equality
help solves many problems where at first we seem to
have not informantion to arrive at the answer .
If a,b,c are positive integers satisfying
`(a)/(b+c)+(b)/(c+a) + (c)/(a+b) = (3)/(2)`, then the value of
`abc + (1)/(abc)`

Options

A. Is `(85)/(4)`
B. Is `(17)/(4)`
C. Is `(65)/(8)`
D. Can't be determined

Answer

To solve the problem, we need to find the value of \( abc + \frac{1}{abc} \) given the condition:

\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{3}{2} \]

where \( a, b, c \) are positive integers.

### Step 1: Analyze the given condition

We start with the equation:

\[ \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} = \frac{3}{2} \]

### Step 2: Test simple values for \( a, b, c \)

Let's first try \( a = b = c = 1 \):

\[ \frac{1}{1+1} + \frac{1}{1+1} + \frac{1}{1+1} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]

This satisfies the equation. Now we calculate \( abc + \frac{1}{abc} \):

\[ abc = 1 \cdot 1 \cdot 1 = 1 \]
\[ abc + \frac{1}{abc} = 1 + \frac{1}{1} = 1 + 1 = 2 \]

### Step 3: Test another set of values

Next, let's try \( a = b = c = 2 \):

\[ \frac{2}{2+2} + \frac{2}{2+2} + \frac{2}{2+2} = \frac{2}{4} + \frac{2}{4} + \frac{2}{4} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]

This also satisfies the equation. Now we calculate \( abc + \frac{1}{abc} \):

\[ abc = 2 \cdot 2 \cdot 2 = 8 \]
\[ abc + \frac{1}{abc} = 8 + \frac{1}{8} = 8 + 0.125 = 8.125 = \frac{65}{8} \]

### Step 4: Test a larger set of values

Now, let's try \( a = b = c = 3 \):

\[ \frac{3}{3+3} + \frac{3}{3+3} + \frac{3}{3+3} = \frac{3}{6} + \frac{3}{6} + \frac{3}{6} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{3}{2} \]

This also satisfies the equation. Now we calculate \( abc + \frac{1}{abc} \):

\[ abc = 3 \cdot 3 \cdot 3 = 27 \]
\[ abc + \frac{1}{abc} = 27 + \frac{1}{27} = 27 + 0.037 = 27.037 = \frac{730}{27} \]

### Conclusion

After testing these values, we find that the maximum value of \( abc + \frac{1}{abc} \) occurs when \( a = b = c = 2 \), yielding:

\[ \frac{65}{8} \]

Thus, the final answer is:

\[ \boxed{\frac{65}{8}} \]