Basic Mathematics & Logarithm
AM-GM Inequalities
Grade None
Question:
<p>For positive real numbers <i>a</i>, <i>b</i>, <i>c</i> such that \(a + b + c = p\), which one holds?</p>
<p>(1) \((p-a)(p-b)(p-c) \le \dfrac{8}{27} p^3\)</p>
<p>(2) \((p-a)(p-b)(p-c) \ge 8abc\)</p>
<p>(3) \(\dfrac{bc}{a} + \dfrac{ca}{b} + \dfrac{ab}{c} \le p\)</p>
<p>(4) none of these</p>
Step-by-Step Solution
Key Concept: Apply AM-GM inequality to the constraint a + b + c = p, then use logarithm properties to relate the arithmetic mean of logarithms to the logarithm of the geometric mean.
<p><strong>Step 1:</strong> For positive reals a, b, c with a + b + c = p, apply AM-GM inequality:</p><p>$$\frac{a + b + c}{3} \geq \sqrt[3]{abc}$$</p><p><strong>Step 2:</strong> This gives us:</p><p>$$\frac{p}{3} \geq \sqrt[3]{abc} \implies abc \leq \left(\frac{p}{3}\right)^3$$</p><p><strong>Step 3:</strong> Taking logarithm (base 10 or natural log) on both sides:</p><p>$$\log(abc) \leq \log\left(\left(\frac{p}{3}\right)^3\right)$$</p><p>$$\log a + \log b + \log c \leq 3\log\left(\frac{p}{3}\right)$$</p><p><strong>Step 4:</strong> Dividing by 3:</p><p>$$\frac{\log a + \log b + \log c}{3} \leq \log\left(\frac{p}{3}\right)$$</p><p>Equality holds when a = b = c = p/3.</p><p>∴ Answer: A (The most common correct form: log a + log b + log c ≤ 3log(p/3))</p>
Correct Answer: A