\( a+b+1\geq \sqrt {ab}+\sqrt {a}+\sqrt {b} \)

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.