# InequalityKnowing Pn(x)=(x-1)(x-2)...(x-n) solve the inequality P1(x)/P2(x)>=2

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We know that Pn(x)=(x-1)(x-2)...(x-n). We have to solve the inequality P1(x)/P2(x)>=2

P1(x) = (x- 1)

P2(x) = (x - 1)(x - 2)

P1(x) / P2(x) = (x- 1)/(x - 1)(x - 2) >=2

=> (x - 1)>=2*(x - 1)(x - 2)

=> (x - 1) - 2*(x - 1)(x - 2) >=0

=> (x - 1)[1 - 2(x - 2)] >=0

=> (x - 1)(5 - 2x) >=0

For this to be true either both (x - 1) and (5 - 2x) >=0 or both are less than 0.

=> x - 1 >=0 and 5 - 2x >=0

=> x >=1 and 5 >=2x

=> x >=1 and 5/2 >=x

So x lies in [1 , 5/2]

x - 1 < 0 and 5 - 2x < 0

=> x < 1 and 5 < 2x

=> x < 1 and 5/2 < x

x cannot be less than 1 and greater than 5/2

**The required values of x are in [1, 5/2]**

If Pn(x)=(x-1)(x-2)...(x-n), then P1(x) = (x-1) and P2(x) = (x-1)(x-2)

P1(x)/P2(x) = (x-1)/(x-1)(x-2)

We'll simplify and we'll get:

P1(x)/P2(x) = 1/(x-2)

But, from enunciation, we know that P1(x)/P2(x) > =2

1/(x-2) >= 2

We'll subtract 2 both sides:

1/(x-2) - 2 >= 0

We'll multiply by (x-2):

(1 - 2x + 4)/(x - 2) > = 0

We'll combine like terms:

(5 - 2x)/(x - 2)>= 0

A ratio is positive if and only if both numerator and denominator, are positive or negative.

Case 1)

5 - 2x >= 0

-2x >= -5

2x =< 5

x = < 5/2 = 2.5

x - 2 >= 0

x >= 2

The interval of admissible values for x, that makes positive the ratio, is [2 ; 2.5].

Case 2)

5 - 2x = < 0

-2x = < -5

2x >= 5

x >= 5/2 = 2.5

x - 2 =< 0

x =< 2

There is no common interval for admissible values for x, in this case.

**The only admissible interval of valid solutions for x is [2 ; 2.5].**