a>0,b>0,a+b=1，求证：（a+1/a)(b+1/b)>25/4

To prove the inequality \(\left(a + \frac{1}{a}\right)\left(b + \frac{1}{b}\right) > \frac{25}{4}\) under the conditions \(a > 0\), \(b > 0\), and \(a + b = 1\), we'll start by using the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step 1: Apply AM-GM inequality
For any positive real number \(x\), the AM-GM inequality states:
\[
\frac{x + \frac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}} = 1
\]
Thus, 
\[
x + \frac{1}{x} \geq 2
\]
Equality holds if and only if \(x = 1\).

### Step 2: Apply the inequality to \(a\) and \(b\)
Using the inequality for \(a\) and \(b\), we have:
\[
a + \frac{1}{a} \geq 2 \quad \text{and} \quad b + \frac{1}{b} \geq 2
\]
So,
\[
\left(a + \frac{1}{a}\right)\left(b + \frac{1}{

Learn how to prove the inequality (a + 1/a)(b + 1/b) > 25/4 given the conditions a > 0, b > 0, and a + b = 1. This problem involves applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Suitable for advanced high school students or anyone interested in inequalities and mathematical proofs.

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