Show that |x|≤2 => (|2x^2 +3x +2|/|x^2 + 2|) ≤8?

|2x² + 3x + 2| / |x² + 2| ≤ 8

|2x² + 3x + 2| ≤ 8|x² + 2|

8|x² + 2| ≥ |2x² + 3x + 2|

|8x² + 16| ≥ |2x² + 3x + 2|

|8x² + 16|² ≥ |2x² + 3x + 2|²

(8x² + 16)² ≥ (2x² + 3x + 2)²

(8x² + 16 + 2x² + 3x + 2)(8x² + 16 - 2x² - 3x - 2) ≥ 0

(10x² + 3x + 18)(6x²- 3x + 14) ≥ 0

10x² + 3x + 18.....D(discriminant) < 0..........10x² + 3x + 18 > 0

6x²- 3x + 14....D(discriminant) < 0....6x²- 3x + 14 > 0

i.e. |2x² + 3x + 2| ≤ 8|x² + 2| is true for all x.

then |x| ≤ 2 => |2x² + 3x + 2| / |x² + 2| ≤ 8　　　　Edit

Start with the following observations: 2x^2 +3x +2 and x^2 + 2 are always positive. So, the absolute value signs are redundant.

(2x^2 +3x +2)/(x^2 + 2) = 2 + (3x - 2)/(x^2 +2)

You you have to show : (3x - 2)/(x^2 +2) < 6 , or equivalently 6x^2 - 3x + 14 >0

But 6x^2 - 3x + 14 is a parabola with a negative discriminate (=9 - 336 ) , hence always positive.

What this shows that (|2x^2 +3x +2|/|x^2 + 2|) ≤8 for all x ( not just |x|≤2).

Suppose that |x| ≤ 2.

Then, |2x² + 3x + 2|/|x² + 2|

= |2x² + 3x + 2|/(x² + 2), since the denominator is positive for all x

≤ |2x² + 3x + 2|/(0 + 2), decreasing the denominator (x² is never negative)

≤ (|2x²| + |3x| + |2|)/2, by the triangle inequality

= (2|x|² + 3|x| + 2)/2

≤ (2 * 2² + 3 * 2 + 2)/2, since |x| ≤ 2

= 8.

Hence, |2x² + 3x + 2|/|x² + 2| ≤ 8.

I hope this helps!