\( (a+b)(a+c)(b+c)\geq 8abc \)

The inequality you've provided, \[ (a + b)(a + c)(b + c) \geq 8abc, \] is a well-known result in algebra, often proven using the **Arithmetic Mean-Geometric Mean (AM-GM) Inequality**. This inequality holds for all **positive real numbers** \( a, b, \) and \( c \). Let's delve into the proof and understand the conditions for equality. ### **Proof Using AM-GM Inequality** The **AM-GM Inequality** states that for any non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. Specifically, for two positive numbers \( x \) and \( y \): \[ \frac{x + y}{2} \geq \sqrt{xy}, \] with equality if and only if \( x = y \). Let's apply this to each pair in your inequality: 1. **First Pair (\( a \) and \( b \)):** \[ a + b \geq 2\sqrt{ab} \] 2. **Second Pair (\( a \) and \( c \)):** \[ a + c \geq 2\sqrt{ac} \] 3. **Third Pair (\( b \) and \( c \)):** \[ b + c \geq 2\sqrt{bc} \] Now, multiply these three inequalities together: \[ (a + b)(a + c)(b + c) \geq (2\sqrt{ab})(2\sqrt{ac})(2\sqrt{bc}) = 8abc \] This directly leads to: \[ (a + b)(a + c)(b + c) \geq 8abc \] ### **Conditions for Equality** The equality in the AM-GM Inequality holds **if and only if** all the paired terms are equal. In the context of your inequality, equality occurs when: \[ a = b = c \] **Example:** Let's verify with \( a = b = c = 1 \): \[ (1 + 1)(1 + 1)(1 + 1) = 2 \times 2 \times 2 = 8 \] \[ 8abc = 8 \times 1 \times 1 \times 1 = 8 \] Both sides are equal, confirming that the equality condition holds when \( a = b = c \). ### **Conclusion** The inequality \[ (a + b)(a + c)(b + c) \geq 8abc \] is always true for positive real numbers \( a, b, \) and \( c \), with equality if and only if \( a = b = c \).