Let `agt0, bgt0, cgt0` and `a+b+c=6` then `((ab+1)^(2))/(b^(2))+((bc+1)^(2))/(c^(2))+((ca+1)^(2))/(a^(2))` may be

Options

A. `75/4`
B. `35`
C. `15`
D. `10`

Explanation

To solve the problem, we need to evaluate the expression

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

given that \( a, b, c > 0 \) and \( a + b + c = 6 \).

### Step 1: Rewrite the expression

We can rewrite each term in the expression as follows:

\[ \frac{(ab + 1)^2}{b^2} = \left(\frac{ab}{b} + \frac{1}{b}\right)^2 = (a + \frac{1}{b})^2 \]

Thus, we can express the entire sum as:

\[ \left(a + \frac{1}{b}\right)^2 + \left(b + \frac{1}{c}\right)^2 + \left(c + \frac{1}{a}\right)^2 \]

### Step 2: Apply the Cauchy-Schwarz Inequality

Using the Cauchy-Schwarz inequality, we can say:

\[ \left( (a + \frac{1}{b})^2 + (b + \frac{1}{c})^2 + (c + \frac{1}{a})^2 \right) \geq \frac{(a+b+c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c})^2}{3} \]

### Step 3: Calculate \( a + b + c \) and \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \)

From the problem, we know \( a + b + c = 6 \).

Now, we need to calculate \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \). By the AM-HM inequality:

\[ \frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \]

This implies:

\[ \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq \frac{9}{6} = \frac{3}{2} \]

### Step 4: Substitute back into the inequality

Now substituting back into our inequality:

\[ \left( (a + \frac{1}{b})^2 + (b + \frac{1}{c})^2 + (c + \frac{1}{a})^2 \right) \geq \frac{(6 + \frac{3}{2})^2}{3} \]

Calculating the right-hand side:

\[ = \frac{(6 + 1.5)^2}{3} = \frac{(7.5)^2}{3} = \frac{56.25}{3} = 18.75 \]

### Step 5: Conclusion

Thus, we have shown that:

\[ \frac{(ab + 1)^2}{b^2} + \frac{(bc + 1)^2}{c^2} + \frac{(ca + 1)^2}{a^2} \geq 18.75 \]

The minimum value of the expression is \( \frac{75}{4} \).