Proof related to inequalities

Question

If $x$ is real then prove that $ \dfrac x {x^2 - 5x + 9}$ lies between $1 $ and $-\frac1{11}$. Actually I am able to prove that the expression's value is $\le 1$ but I am not able to prove it to be $\ge -\frac1{11}$ by using the quadratic formula. Can someone help? Thanks in advance.

Answer 1

Let $$k = \frac{x}{x^2-5x+9}$$

$$kx^2+(-5k-1)x+9k=0$$

We required $$(-5k-1)^2-4k(9k) \ge 0$$

$$(5k+1)^2-(6k)^2 \ge 0$$

You just have to solve this inequality to find the valid range.