Proove that if x is a positive real number then [(x+1)/(x+2)]<[(x+3)/(x+4)]?

Please read :

[(x + 1)/(x + 2)] < [(x + 3)/(x + 4)] → if it's really the case, continue

[(x + 1)/(x + 2)] < [(x + 3)/(x + 4)] → do not make the cross multiply, put all the items on the left side

[(x + 1)/(x + 2)] - [(x + 3)/(x + 4)] < 0 → common denominator

[(x + 1).(x + 4)/(x + 2).(x + 4)] - [(x + 3).(x + 2)/(x + 2).(x + 4)] < 0

[(x + 1).(x + 4) - (x + 3).(x + 2)] / [(x + 2).(x + 4)] < 0

[(x² + 4x + x + 4) - (x² + 2x + 3x + 6)] / [(x + 2).(x + 4)] < 0

[(x² + 5x + 4) - (x² + 5x + 6)] / [(x + 2).(x + 4)] < 0

(x² + 5x + 4 - x² - 5x - 6) / [(x + 2).(x + 4)] < 0

- 2 / [(x + 2).(x + 4)] < 0 → you multiply both sides by (- 1), as (- 1) is negative, you change the direction

{ - 2 / [(x + 2).(x + 4)] } * (- 1) > 0 * (- 1) → you simplify

2 / [(x + 2).(x + 4)] > 0


Even if this polynomial is a fraction, its sign will be the same that the product, i.e.:

2.(x + 2).(x + 4) > 0 → you know that 2 is always > 0

(x + 2).(x + 4) > 0


You said that x is a positive real number, so you can write: (x + 2) > 0

You said that x is a positive real number, so you can write: (x + 4) > 0

The sign of the product of two positive factors gives you a positive number.


Conclusion:

If x is a positive real number then: (x + 2).(x + 4) > 0

→ so you can conclude that: [(x + 1)/(x + 2)] < [(x + 3)/(x + 4)]

Given that x is positive, you can multiply both sides by x+2 and not affect the direction of the inequality symbol. That's because x+2 will also be positive.

x + 1 < (x + 2)(x + 3) / (x + 4)

Next multiply both sides by x+4, which again is positive:
(x + 1)(x + 4) < (x + 2)(x + 3)

Expand that out:
x² + 5x + 5 < x² + 5x + 6

We can subtract the same value from both sides and not affect the inequality, so subtract x² and 5x from both sides:
5 < 6

We have gotten to a true statement so that shows our original inequality must have been true.