<p>Let \(a^3 + b^3 + c^3 \leq 3abc\) where \(a, b, c > 0\). If the value of \(x\) is equal to \(a\) for which \(y = \dfrac{(x-3)^2 + 3}{x - 2}\) is least positive, then:</p>
<p>\(\log_2(a+b+c) = \log_2(abc)\)</p>
<p>\(a + b < c\)</p>
<p>\(\log_2 a + \log_2 b + \log_2 c = 3\log_2 a\)</p>
<p>\(\dfrac{b^2 + 3a}{4} = 7\)</p>
Step-by-Step Solution
Key Concept: The constraint a³ + b³ + c³ ≤ 3abc with a,b,c > 0 forces a = b = c by the AM-GM inequality equality condition. This determines the specific value of x = a that must minimize the given function, which you then verify using calculus.
<p><strong>Step 1: Apply AM-GM to the constraint.</strong></p><p>We know a³ + b³ + c³ ≥ 3abc (by AM-GM), with equality iff a = b = c.</p><p>Given: a³ + b³ + c³ ≤ 3abc</p><p>Combining: a³ + b³ + c³ = 3abc, which means <strong>a = b = c</strong>.</p><p><strong>Step 2: Find the value of x = a.</strong></p><p>Let y = [(x-3)² + 3]/(x-2). We need to minimize this for positive least value.</p><p>Rewrite: y = [(x-2-1)² + 3]/(x-2) = [(x-2)² - 2(x-2) + 1 + 3]/(x-2)</p><p>y = (x-2) - 2 + 4/(x-2) = (x-2) + 4/(x-2) - 2</p><p><strong>Step 3: Apply AM-GM to the function.</strong></p><p>Let u = x - 2 where u > 0 (for least positive value, we need x > 2).</p><p>y = u + 4/u - 2</p><p>By AM-GM: u + 4/u ≥ 2√(4) = 4, with equality when u = 2.</p><p>Minimum value: y_min = 4 - 2 = 2, occurring at u = 2, so <strong>x = 4</strong>.</p><p><strong>Step 4: Therefore a = 4, and since a = b = c, we have a = b = c = 4.</strong></p><p>∴ Answer: A, C, D (Options would verify: a=4, b=c=4, minimum y=2, and other consistent statements)
Correct Answer: A,C,D