Prove: If x>1, then (x-1)(x^2+3x+1) > 0?

Answer 1

If x > 1, then x-1 > 0

Now let's check the roots of the quadratic expression: -1 and -2

So, the sign is this way: positive in (-oo, -2) and (-1, + oo), 0 if x = -2 or -1 and negative in the rest of R

So, if x > 1 both factors are positive and so is their product.



Ilusion

Answer 2

since x>1, we let x=2, then

(2-1)(2^2+3(2)+1) > 0
(1)(3+9+1) > 0
(1)(13) > 0
13 > 0
thus, statement is true for x=2 as so for numbers greater than 1..

Answer 3

(x-1)(x^2+3x+1)
Notice that if x > 1, then x^2 > 1, obviously 3x > 1
Therefore, x^2 + 3x + 1 > 3
And x > 1, then x - 1 > 0
Hence,
(x-1)(x^2+3x+1) > 0

Answer 4

If x > 1 then both (x - 1) and (x² + 3x + 1) are positive and their product must be > 0

QED

Doug

Answer 5

(x-1)(x^2+3x+1) > 0
x^3 + 3x^2 + x - x^2 - 3x -1 > 0
x^3 +2x^2 -2x -1 > 0
x^3 +2x^2 > 2x + 1
x^2 + 2x > 2 + 1/x

a) For x -> 1+

LHS: lim x-> 1+ (x^2 + 2x) = 3+
RHS: lim x->1+ (2 + 1/x) = 3-

b) For x -> oo
LHS: lim x->oo (x^2 + 2x) -> oo
RHS:lim x->oo( 2 + 1/x) = 2 + lim x->oo(1/x) = 2+

Since LHS is monotonously rising and RHS is monotonously falling, the inequality holds true for x > 1.

Answer 6

You first.