$$ a+b+1\geq \sqrt {ab}+\sqrt {a}+\sqrt {b} $$

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To analyze and prove the inequality:

$$
a + b + 1 \geq \sqrt{ab} + \sqrt{a} + \sqrt{b}
$$

we'll proceed step-by-step. Let's assume that $$ a $$ and $$ b $$ are non-negative real numbers ($$ a, b \geq 0 $$) to ensure that the square roots are defined.

### Step 1: Variable Substitution

Let’s introduce new variables to simplify the inequality:

$$
x = \sqrt{a} \quad \text{and} \quad y = \sqrt{b}
$$

Since $$ a, b \geq 0 $$, it follows that $$ x, y \geq 0 $$. Substituting these into the original inequality:

$$
a = x^2 \quad \text{and} \quad b = y^2
$$

The inequality becomes:

$$
x^2 + y^2 + 1 \geq xy + x + y
$$

### Step 2: Rearranging the Inequality

Rearrange the inequality to group like terms:

$$
x^2 - xy + y^2 - x - y + 1 \geq 0
$$

### Step 3: Analyzing as a Quadratic in $$ x $$

Consider the expression as a quadratic in $$ x $$:

$$
x^2 - (y + 1)x + (y^2 - y + 1) \geq 0
$$

For this quadratic to be non-negative for all $$ x $$, its discriminant must be less than or equal to zero.

### Step 4: Calculating the Discriminant

The discriminant $$ D $$ of the quadratic $$ ax^2 + bx + c $$ is given by $$ D = b^2 - 4ac $$.

For our quadratic:

$$
a = 1, \quad b = -(y + 1), \quad c = y^2 - y + 1
$$

Thus,

$$
D = (y + 1)^2 - 4(1)(y^2 - y + 1) = y^2 + 2y + 1 - 4y^2 + 4y - 4 = -3y^2 + 6y - 3
$$

Factor out the negative:

$$
D = -3(y^2 - 2y + 1) = -3(y - 1)^2
$$

Since $$ (y - 1)^2 \geq 0 $$ for all real $$ y $$, it follows that:

$$
D \leq 0
$$

### Step 5: Conclusion from the Discriminant

A non-positive discriminant ($$ D \leq 0 $$) means that the quadratic $$ x^2 - (y + 1)x + (y^2 - y + 1) $$ does not cross the x-axis and is always non-negative (since the coefficient of $$ x^2 $$ is positive).

### Step 6: Finalizing the Proof

Since the rearranged inequality holds for all $$ x, y \geq 0 $$, this implies that the original inequality:

$$
a + b + 1 \geq \sqrt{ab} + \sqrt{a} + \sqrt{b}
$$

is true for all non-negative real numbers $$ a $$ and $$ b $$.

### Example Verification

Let's verify the inequality with specific values:

- **Case 1:** $$ a = b = 1 $$
  
  $$
  1 + 1 + 1 = 3 \geq \sqrt{1 \cdot 1} + \sqrt{1} + \sqrt{1} = 1 + 1 + 1 = 3
  $$
  
  The inequality holds with equality.

- **Case 2:** $$ a = 4, b = 9 $$
  
  $$
  4 + 9 + 1 = 14 \geq \sqrt{4 \cdot 9} + \sqrt{4} + \sqrt{9} = 6 + 2 + 3 = 11
  $$
  
  $$ 14 \geq 11 $$ holds true.

- **Case 3:** $$ a = 0, b = 0 $$
  
  $$
  0 + 0 + 1 = 1 \geq \sqrt{0 \cdot 0} + \sqrt{0} + \sqrt{0} = 0 + 0 + 0 = 0
  $$
  
  $$ 1 \geq 0 $$ holds true.

These examples confirm that the inequality holds in various scenarios.
The inequality $$ a + b + 1 \geq \sqrt{ab} + \sqrt{a} + \sqrt{b} $$ holds true for all non-negative real numbers $$ a $$ and $$ b $$.

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