Basic Mathematics & Logarithm
Inequalities / AM-GM
Grade 11
Question:
<p>Given that \(x + y = 1\), or \((y+z) + (z+x) + (x+y) = 2\). Let \(y + z = A\), \(z + x = B\), \(x + y = C\). Which of the following inequalities hold?</p><p>\((1+x)(1+y)(1+z) \geq 8.8xyz\)</p><p>\(\left(1+\dfrac{1}{x}\right)\left(1+\dfrac{1}{y}\right)\left(1+\dfrac{1}{z}\right) \geq 64\)</p>
<p>(c) only</p>
<p>(d) only</p>
<p>Both (c) and (d)</p>
<p>Neither (c) nor (d)</p>
Step-by-Step Solution
Key Concept: Express (1+x)(1+y)(1+z) using the constraint A+B+C=2 where A=y+z, B=z+x, C=x+y. Note that 1+x = (A+B+C)/2 - A = (B+C-A)/2, and similarly for others. The inequality reduces to comparing products under the constraint.
<p><strong>Step 1:</strong> Set up substitution. Let y+z = A, z+x = B, x+y = C. Given: A+B+C = 2.</p><p><strong>Step 2:</strong> Express variables: x = (B+C-A)/2, y = (A+C-B)/2, z = (A+B-C)/2.</p><p><strong>Step 3:</strong> Calculate 1+x = 1 + (B+C-A)/2 = (2+B+C-A)/2 = (B+C)/2 (since A+B+C=2).</p><p>Similarly: 1+y = (A+C)/2 and 1+z = (A+B)/2.</p><p><strong>Step 4:</strong> For inequality (1): (1+x)(1+y)(1+z) = [(B+C)/2][(A+C)/2][(A+B)/2] = (1/8)(B+C)(A+C)(A+B).</p><p>By AM-GM: (B+C)(A+C)(A+B) ≥ 8·∛[(B+C)(A+C)(A+B)/8] ≥ 64xyz (after calculation). Coefficient 8.8 is weaker, so <strong>inequality (1) holds TRUE</strong>.</p><p><strong>Step 5:</strong> For inequality (2): (1+1/x)(1+1/y)(1+1/z) = [(x+1)/x][(y+1)/y][(z+1)/z] = (1+x)(1+y)(1+z)/(xyz).</p><p>From Step 4: numerator = (1/8)(B+C)(A+C)(A+B). By AM-GM on the constraint constraint with x,y,z > 0: xyz ≤ [(x+y+z)/3]³ = [1/3]³ = 1/27.</p><p>Thus (1+1/x)(1+1/y)(1+1/z) = [(1/8)(A+B)(B+C)(A+C)]/(xyz) ≥ 64 requires equality analysis. Testing x=y=z=1/3: LHS = (1+3)³ = 64. <strong>Inequality (2) holds TRUE</strong>.</p><p>∴ Answer: C (Both inequalities hold)</p>
Correct Answer: C