If `a ,b ,c in R^+, t h e n(b c)/(b+c)+(a c)/(a+c)+(a b)/(a+b)` is always

Options

A. `le(1)/(2)(a+b+c)`
B. `ge (1)/(2) sqrt(abc)`
C. `le(1)/(3) (a+b+c)`
D. `ge(1)/(3)sqrt(abc)`

Answer

To solve the given problem, we need to show that

\[ \frac{bc}{b+c} + \frac{ac}{a+c} + \frac{ab}{a+b} \leq \frac{1}{2}(a + b + c) \]

for all positive real numbers \(a\), \(b\), and \(c\).

### Step 1: Understand the Inequality
We will use the concept of the Arithmetic Mean-Harmonic Mean (AM-HM) inequality. The AM-HM inequality states that for any positive real numbers \(x\) and \(y\):

\[ \frac{x+y}{2} \geq \frac{2xy}{x+y} \]

### Step 2: Apply AM-HM Inequality
We can apply the AM-HM inequality to each of the three pairs \((a, b)\), \((b, c)\), and \((c, a)\):

1. For \(a\) and \(b\):
\[ \frac{a+b}{2} \geq \frac{2ab}{a+b} \implies \frac{ab}{a+b} \leq \frac{a+b}{4} \]

2. For \(b\) and \(c\):
\[ \frac{b+c}{2} \geq \frac{2bc}{b+c} \implies \frac{bc}{b+c} \leq \frac{b+c}{4} \]

3. For \(c\) and \(a\):
\[ \frac{c+a}{2} \geq \frac{2ac}{c+a} \implies \frac{ac}{a+c} \leq \frac{c+a}{4} \]

### Step 3: Combine the Results
Now, we can combine these inequalities:

\[ \frac{ab}{a+b} + \frac{bc}{b+c} + \frac{ac}{a+c} \leq \frac{a+b}{4} + \frac{b+c}{4} + \frac{c+a}{4} \]

### Step 4: Simplify the Right Side
The right side simplifies as follows:

\[ \frac{a+b}{4} + \frac{b+c}{4} + \frac{c+a}{4} = \frac{(a+b) + (b+c) + (c+a)}{4} = \frac{2(a+b+c)}{4} = \frac{a+b+c}{2} \]

### Step 5: Conclusion
Thus, we have shown that:

\[ \frac{bc}{b+c} + \frac{ac}{a+c} + \frac{ab}{a+b} \leq \frac{a+b+c}{2} \]

This completes the proof that the given expression is always less than or equal to \(\frac{1}{2}(a + b + c)\).