If positive numbers a, b, c are in H.P, then minimum value of `(a+b)/(2a-b)+(c+b)/(2c-b)` is

Options

A. `1
B. 2
C. 3
D. 4

Answer

To find the minimum value of the expression \(\frac{a+b}{2a-b} + \frac{c+b}{2c-b}\) given that \(a\), \(b\), and \(c\) are in Harmonic Progression (H.P.), we can follow these steps:

### Step 1: Understand the condition of H.P.
If \(a\), \(b\), and \(c\) are in H.P., then their reciprocals \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P. This means:
\[ \frac{1}{b} - \frac{1}{a} = \frac{1}{c} - \frac{1}{b} \]
This can be rearranged to:
\[ \frac{2}{b} = \frac{1}{a} + \frac{1}{c} \]

### Step 2: Substitute and simplify the expression
Using the relationship from H.P., we can express \(b\) in terms of \(a\) and \(c\):
\[ b = \frac{2ac}{a+c} \]
Now substitute \(b\) into the original expression:
\[ \frac{a + \frac{2ac}{a+c}}{2a - \frac{2ac}{a+c}} + \frac{c + \frac{2ac}{a+c}}{2c - \frac{2ac}{a+c}} \]

### Step 3: Simplify each term
Let's simplify the first term:
\[ \frac{a + \frac{2ac}{a+c}}{2a - \frac{2ac}{a+c}} = \frac{\frac{a(a+c) + 2ac}{a+c}}{\frac{(2a(a+c) - 2ac)}{a+c}} = \frac{a^2 + ac + 2ac}{2a^2 + 2ac - 2ac} = \frac{a^2 + 3ac}{2a^2} \]
Now simplify the second term similarly:
\[ \frac{c + \frac{2ac}{a+c}}{2c - \frac{2ac}{a+c}} = \frac{c(a+c) + 2ac}{2c(a+c) - 2ac} = \frac{c^2 + 3ac}{2c^2} \]

### Step 4: Combine the simplified terms
Now we have:
\[ \frac{a^2 + 3ac}{2a^2} + \frac{c^2 + 3ac}{2c^2} \]
To find a common denominator:
\[ = \frac{(a^2 + 3ac)c^2 + (c^2 + 3ac)a^2}{2a^2c^2} \]

### Step 5: Analyze the expression
To find the minimum value, we can use the Cauchy-Schwarz inequality:
\[ \left( \frac{a^2 + 3ac}{2a^2} + \frac{c^2 + 3ac}{2c^2} \right) \geq \frac{(1 + 1)^2}{2 + 2} = 2 \]
Thus, the minimum value of the expression \(\frac{a+b}{2a-b} + \frac{c+b}{2c-b}\) is 4.

### Conclusion
The minimum value of \(\frac{a+b}{2a-b} + \frac{c+b}{2c-b}\) is \(4\).

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