<p>If <i>G</i> and <i>L</i> are the greatest and least values of the expression \(\frac{x^2 - x + 1}{x^2 + x + 1}\), \(x \in \mathbb{R}\) respectively, then the least value of \(G^5 + L^5\) is</p>
Step-by-Step Solution
Key Concept: Find the range of the rational expression using the discriminant condition for real x, then apply AM-GM inequality.
<p><strong>Solution:</strong></p><p>Let $y = \frac{x^2 - x + 1}{x^2 + x + 1}$</p><p>Then $xy + xy + y = x^2 - x + 1$</p><p>$(y - 1)x^2 + (y + 1)x + y - 1 = 0$</p><p>For $x \in \mathbb{R}$, the discriminant must be $\geq 0$:</p><p>$(y + 1)^2 - 4(y - 1)(y - 1) \geq 0$</p><p>$(y + 1)^2 - (2y - 2)^2 \geq 0$</p><p>$(3y - 1)(y - 3) \leq 0$</p><p>$\frac{1}{3} \leq y \leq 3$</p><p>Therefore, $G = 3$ and $L = \frac{1}{3}$</p><p>$GL = 1$, so $(GL)^{1/5} = 1$</p><p>$G^5 + L^5 \geq 2(GL) = 2$</p><p>The minimum value of $G^5 + L^5$ is <strong>2</strong>.</p>
Correct Answer: b