a, b are positive real numbers such that `1/a+9/b=1` The smallest value of `a + b ` is

Answer

To find the smallest value of \( a + b \) given the equation \( \frac{1}{a} + \frac{9}{b} = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
From the equation \( \frac{1}{a} + \frac{9}{b} = 1 \), we can express \( b \) in terms of \( a \):
\[ \frac{9}{b} = 1 - \frac{1}{a} \]
This simplifies to:
\[ b = \frac{9a}{a - 1} \]

### Step 2: Express \( a + b \)
Now we can substitute \( b \) into the expression \( a + b \):
\[ a + b = a + \frac{9a}{a - 1} \]
To combine these, we find a common denominator:
\[ a + b = \frac{a(a - 1) + 9a}{a - 1} = \frac{a^2 - a + 9a}{a - 1} = \frac{a^2 + 8a}{a - 1} \]

### Step 3: Minimize \( a + b \)
To minimize \( a + b \), we can use calculus or the AM-GM inequality. Here, we will apply the AM-GM inequality.

Let \( x = a - 1 \), then \( a = x + 1 \). We substitute this back into our expression:
\[ a + b = \frac{(x + 1)^2 + 8(x + 1)}{x} = \frac{x^2 + 2x + 1 + 8x + 8}{x} = \frac{x^2 + 10x + 9}{x} \]
This simplifies to:
\[ a + b = x + 10 + \frac{9}{x} \]

### Step 4: Apply AM-GM Inequality
By the AM-GM inequality:
\[ x + \frac{9}{x} \geq 2\sqrt{x \cdot \frac{9}{x}} = 6 \]
Thus, we have:
\[ a + b \geq 10 + 6 = 16 \]

### Step 5: Conclusion
The minimum value of \( a + b \) is \( 16 \). This occurs when \( x = 3 \) (which means \( a - 1 = 3 \) or \( a = 4 \)), and substituting back gives \( b = 9 \).

### Final Answer
The smallest value of \( a + b \) is \( \boxed{16} \).

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