Find the minimum value of `(a+1/a)^(2) +(b+1/b)^(2)` where `a gt 0, b gt 0 ` and `a+b = sqrt(15)`

Answer

To find the minimum value of \( (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \) given that \( a > 0, b > 0 \) and \( a + b = \sqrt{15} \), we can follow these steps:

### Step 1: Use the AM-GM Inequality
We know that for any positive \( x \), the inequality \( x + \frac{1}{x} \geq 2 \) holds. Thus, we can apply this to both \( a \) and \( b \):

\[ a + \frac{1}{a} \geq 2 \quad \text{and} \quad b + \frac{1}{b} \geq 2 \]

### Step 2: Square the Inequalities
Squaring both sides gives us:

\[ (a + \frac{1}{a})^2 \geq 4 \quad \text{and} \quad (b + \frac{1}{b})^2 \geq 4 \]

### Step 3: Sum the Inequalities
Adding these inequalities together, we have:

\[ (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geq 4 + 4 = 8 \]

### Step 4: Apply the Cauchy-Schwarz Inequality
To find a more precise lower bound, we can apply the Cauchy-Schwarz inequality:

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

Given \( a + b = \sqrt{15} \), we need to find \( \frac{1}{a} + \frac{1}{b} \).

### Step 5: Express \( \frac{1}{a} + \frac{1}{b} \)
Using the identity \( \frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} \), we can express this in terms of \( a + b \) and \( ab \).

### Step 6: Find \( ab \) using \( a + b \)
Let \( ab = k \). By the AM-GM inequality, we have:

\[ \frac{a + b}{2} \geq \sqrt{ab} \Rightarrow \frac{\sqrt{15}}{2} \geq \sqrt{k} \Rightarrow k \leq \frac{15}{4} \]

### Step 7: Substitute back into the Cauchy-Schwarz Inequality
Now we can substitute back into our earlier inequality:

\[ \sqrt{15}\left((a + \frac{1}{a})^2 + (b + \frac{1}{b})^2\right) \geq \left(\sqrt{15} + \frac{1}{a} + \frac{1}{b}\right)^2 \]

### Step 8: Minimize the Expression
To minimize \( (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \), we can set \( a = b \). Therefore, \( 2a = \sqrt{15} \Rightarrow a = b = \frac{\sqrt{15}}{2} \).

### Step 9: Calculate the Minimum Value
Substituting \( a = b = \frac{\sqrt{15}}{2} \) into the expression:

\[ (a + \frac{1}{a})^2 = \left(\frac{\sqrt{15}}{2} + \frac{2}{\sqrt{15}}\right)^2 = \left(\frac{\sqrt{15}}{2} + \frac{2\sqrt{15}}{15}\right)^2 = \left(\frac{\sqrt{15}}{2} + \frac{2\sqrt{15}}{15}\right)^2 \]

Calculating this gives:

\[ = \left(\frac{15 + 4}{2\sqrt{15}}\right)^2 = \left(\frac{19}{2\sqrt{15}}\right)^2 = \frac{361}{60} \]

Thus, the minimum value of \( (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \) is:

\[ \frac{361}{30} \]

### Final Answer
The minimum value of \( (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \) is \( \frac{361}{30} \).