Let `x^2+y^2+x y+1geqa(x+y)AAx ,y in R ,` then the number of possible integer (s) in the range of `a` is___________.

Explanation

To solve the inequality \( x^2 + y^2 + xy + 1 \geq a(x + y) \) where \( x, y \in \mathbb{R} \), we can follow these steps:

### Step 1: Rearranging the Inequality
We start with the given inequality:
\[ x^2 + y^2 + xy + 1 \geq a(x + y) \]
Rearranging this gives:
\[ x^2 + y^2 + xy + 1 - ax - ay \geq 0 \]
This can be rewritten as:
\[ x^2 + y^2 + xy - ax - ay + 1 \geq 0 \]

### Step 2: Grouping Terms
We can group the terms involving \( x \) and \( y \):
\[ x^2 + (y - a)x + y^2 - ay + 1 \geq 0 \]
This is a quadratic in \( x \):
\[ x^2 + (y - a)x + (y^2 - ay + 1) \geq 0 \]

### Step 3: Discriminant Condition
For the quadratic in \( x \) to be non-negative for all \( x \), the discriminant must be less than or equal to zero:
\[ D = (y - a)^2 - 4(y^2 - ay + 1) \leq 0 \]
Expanding this:
\[ D = (y - a)^2 - 4y^2 + 4ay - 4 \]
Simplifying gives:
\[ D = y^2 - 2ay + a^2 - 4y^2 + 4ay - 4 \leq 0 \]
\[ D = -3y^2 + 2ay + (a^2 - 4) \leq 0 \]

### Step 4: Another Quadratic in \( y \)
We can rearrange this into a standard quadratic form:
\[ 3y^2 - 2ay - (a^2 - 4) \geq 0 \]
For this quadratic to be non-negative for all \( y \), its discriminant must also be less than or equal to zero:
\[ D' = (-2a)^2 - 4 \cdot 3 \cdot (-(a^2 - 4)) \leq 0 \]
Calculating the discriminant:
\[ D' = 4a^2 + 12(a^2 - 4) \leq 0 \]
\[ D' = 4a^2 + 12a^2 - 48 \leq 0 \]
\[ 16a^2 - 48 \leq 0 \]
\[ 16a^2 \leq 48 \]
\[ a^2 \leq 3 \]

### Step 5: Finding the Range of \( a \)
Taking the square root gives:
\[ -\sqrt{3} \leq a \leq \sqrt{3} \]

### Step 6: Identifying Possible Integer Values
The integers in the range \( [-\sqrt{3}, \sqrt{3}] \) are:
- \( -1 \)
- \( 0 \)
- \( 1 \)

Thus, the number of possible integers in the range of \( a \) is:
\[ \text{Number of integers} = 3 \]

### Final Answer
The number of possible integers in the range of \( a \) is **3**.

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