If a + b =1, a `gt` 0,b `gt` 0, prove that `(a + (1)/(a))^(2) + (b + (1)/(b))^(2) ge (25)/(2)`

Explanation

To prove that \( (a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 \geq \frac{25}{2} \) given that \( a + b = 1 \) and \( a > 0, b > 0 \), we can follow these steps:

### Step 1: Use the Cauchy-Schwarz Inequality
We start by applying the Cauchy-Schwarz inequality which states that for any non-negative real numbers \( x_1, x_2, y_1, y_2 \):
\[ (x_1^2 + x_2^2)(y_1^2 + y_2^2) \geq (x_1y_1 + x_2y_2)^2 \]
Let \( x_1 = a + \frac{1}{a} \) and \( x_2 = b + \frac{1}{b} \). Then we can express our inequality as:
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