We call a a good number, if the inequality $\frac{2 x^{2} + 2 x + 3}{x^{2} + x + 1} ⩽ a$ is satisfied for any real $x$ .
(a) Prove that $4$ is a good number.
(b) Find all good numbers.

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Solution

$\frac{2 x^{2} + 2 x + 3}{x^{2} + x + 1} \leq 4$
$2 x^{2} + 2 x + 3 \leq 4 x^{2} + 4 x + 4$
$2 x^{2} + 2 x + 1 \geq 0$
As $△ < 0$ so $2 x^{2} + 2 x + 1 \geq 0$ for all
$x \in R$
$\frac{2 x^{2} + 2 x + 3}{x^{2} + x + 1} \leq a$
$2 x^{2} + 2 x + 3 \leq a x^{2} + a x + a$
$(a - 2) x^{2} + (a - 2) x + (a - 3) \geq 0$ for all $x \in R$
$(a - 2)^{2} - 4 (a - 2) (a - 3) \leq 0$
$a^{2} - 4 a + 4 - 4 (a^{2} - 5 a + 6) \leq 0$
$a^{2} - 4 a + 4 - 4 a^{2} + 20 a - 24 \leq 0$
$- 3 a^{2} + 16 a - 20 \leq 0$
$3 a^{2} - 16 a - 10 a + 20 \geq 0$
$3 a^{2} - 6 a - 10 a + 20 \geq 0$
$(3 a - 10) (a - 2) \geq 0$
$a \in (- \infty, 2) \cup (\frac{10}{3}, \infty)$
Since $4$ lies in the satisfies the inequality, $4$ is a good number.

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