`f:RrarrR` be twice differentiable function satisfying
`f^(")(x)-5f^(')(x)+6f(x)ge0AAxge0` if `f(0)=1`
`f^(')(0)=0`. If `f(x)` satisfies `f(x),f(x)geae^(bx)-be^(ax), Aaxge0`, then find `(a+b)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given inequality
We start with the inequality:
\[ f''(x) - 5f'(x) + 6f(x) \geq 0 \quad \text{for } x \geq 0 \]
This can be rewritten as:
\[ f''(x) - 3f'(x) - 2f(x) \geq 0 \]

### Step 2: Define a new function
Let us define:
\[ g(x) = f'(x) - 2f(x) \]
Then, we can express \(f''(x)\) in terms of \(g(x)\):
\[ f''(x) = g'(x) + 2f'(x) \]
Substituting this into our inequality gives:
\[ g'(x) + 2f'(x) - 3f'(x) - 2f(x) \geq 0 \]
Which simplifies to:
\[ g'(x) - f'(x) - 2f(x) \geq 0 \]

### Step 3: Rearranging the inequality
We can rearrange the inequality:
\[ g'(x) - 3g(x) \geq 0 \]
This implies:
\[ g'(x) \geq 3g(x) \]

### Step 4: Analyze the function \(g(x)\)
The inequality \(g'(x) - 3g(x) \geq 0\) suggests that \(g(x)e^{-3x}\) is a non-decreasing function. Thus, we can write:
\[ g(x)e^{-3x} \geq g(0)e^{-3 \cdot 0} = g(0) \]

### Step 5: Calculate \(g(0)\)
Now, we need to find \(g(0)\):
\[ g(0) = f'(0) - 2f(0) = 0 - 2 \cdot 1 = -2 \]
Thus, we have:
\[ g(x)e^{-3x} \geq -2 \]

### Step 6: Express \(f(x)\)
From \(g(x) = f'(x) - 2f(x)\), we can express \(f'(x)\):
\[ f'(x) = g(x) + 2f(x) \]

### Step 7: Solve for \(f(x)\)
We can find a lower bound for \(f(x)\) based on the derived inequalities. We know:
\[ f(x)e^{-2x} \geq -2e^{3x} \]
This implies:
\[ f(x) \geq 3e^{2x} - 2e^{3x} \]

### Step 8: Compare with given function
The given condition states:
\[ f(x) \geq ae^{bx} - be^{ax} \]
By comparing, we can identify:
- \(a = 3\)
- \(b = 2\)

### Step 9: Find \(a + b\)
Finally, we find:
\[ a + b = 3 + 2 = 5 \]

Thus, the final answer is:
\[ \boxed{5} \]