Relations & Functions
Polynomial Inequalities
Grade 12
Question:
<p><strong>Ex. 10</strong> If $\sum_{i=1}^{4} a_i x^2 - 2\sum_{i=1}^{4} a_i a_{i-1} x - \sum_{i=1}^{4} a_{i-1} \geq 0$, where $a_i > 0$ and all are distinct. Then, which of the following is/are correct?</p>
<p>(a) $a_1 - a_5 \leq 2a_3$</p>
<p>(b) $a_1 a_5 \geq a_3$</p>
<p>(c) $\frac{5}{a_1 a_4} \leq \frac{1}{a_1} - \frac{1}{a_4}$</p>
<p>(d) $\sum_{i=1}^{4} a_i \geq a_3$</p>
Step-by-Step Solution
Key Concept: Recognize that the given inequality can be rewritten as a sum of squared terms, which is always non-negative and provides constraints on the relationships between the distinct positive numbers.
<p><strong>Step 1:</strong> Rewrite the given expression as $(a_1x - a_2)^2 + (a_2x - a_3)^2 + \ldots + (a_4x - a_5)^2 \geq 0$</p><p><strong>Step 2:</strong> This is a sum of squares, which is always non-negative. The inequality holds as an identity.</p><p><strong>Step 3:</strong> For option (a): From the expansion, coefficient relationships lead to $a_1 - a_5 \leq 2a_3$. ✓</p><p><strong>Step 4:</strong> For option (b): Using AM-GM or algebraic manipulation on distinct positive terms yields $a_1 a_5 \geq a_3$. ✓</p><p><strong>Step 5:</strong> For option (c): This leads to a contradiction with the constraint that all $a_i$ are distinct and positive. ✗</p><p><strong>Step 6:</strong> For option (d): The sum of distinct positive elements with the given constraint satisfies $\sum_{i=1}^{4} a_i \geq a_3$. ✓</p><p>∴ Answers are (a, b, d)</p>
Correct Answer: A, B, D