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

To prove the inequality \( (a+b)(a+c)(b+c) \geq 8abc \), we can use the AM-GM (Arithmetic Mean-Geometric Mean) inequality. ### Step 1: Apply AM-GM Inequality According to the AM-GM inequality, for any non-negative real numbers \( x_1, x_2, \ldots, x_n \): \[ \frac{x_1 + x_2 + \ldots + x_n}{n} \geq \sqrt[n]{x_1 x_2 \ldots x_n} \] ### Step 2: Set Up the Terms We can apply AM-GM to the three pairs of terms in our expression: 1. \( a+b \) 2. \( a+c \) 3. \( b+c \) ### Step 3: Apply AM-GM to Each Pair Using AM-GM on each pair, we have: \[ \frac{(a+b) + (a+c) + (b+c)}{3} \geq \sqrt[3]{(a+b)(a+c)(b+c)} \] Calculating the left side: \[ (a+b) + (a+c) + (b+c) = 2a + 2b + 2c = 2(a+b+c) \] Thus, we can rewrite the inequality as: \[ \frac{2(a+b+c)}{3} \geq \sqrt[3]{(a+b)(a+c)(b+c)} \] ### Step 4: Rearranging the Inequality Cubing both sides gives: \[ \left(\frac{2(a+b+c)}{3}\right)^3 \geq (a+b)(a+c)(b+c) \] ### Step 5: Expand the Left Side Calculating the left side: \[ \frac{8(a+b+c)^3}{27} \] ### Step 6: Expand the Right Side Now we need to show that: \[ \frac{8(a+b+c)^3}{27} \geq 8abc \] Dividing both sides by 8 gives: \[ \frac{(a+b+c)^3}{27} \geq abc \] ### Step 7: Apply AM-GM Again Now, we can apply AM-GM to \( a, b, c \): \[ \frac{a+b+c}{3} \geq \sqrt[3]{abc} \] Cubing both sides gives: \[ \frac{(a+b+c)^3}{27} \geq abc \] ### Conclusion Combining these results, we have shown that: \[ (a+b)(a+c)(b+c) \geq 8abc \] Thus, the inequality \( (a+b)(a+c)(b+c) \geq 8abc \) holds true for all non-negative \( a, b, c \).