inequalities help (1 Viewer)

[ctpengage]

prove that the inequality x(x+1)(x+2)(x+3) >= -1

holds for all real numbers x

 

[Just.Snaz]

well, I did it proving by inductions
step 1, you sub n = 1 it'll obviously be <= -1
step 2, assume true for n = k
k(k+1)(k+2)(k+3) >= -1
and prove true for n = k+1
ie, LHS = (k+1)(K+2)(k+3)(k+4)
>= (-1/k)(k+4) [from assumption]
and so even if you minus something, the LHS is still greater either way
so..
>= -1

I think that's right..

EDIT: i think the -1/k is a bit controversial so better wait for another reply

 

[tommykins]

well, I did it proving by inductions
step 1, you sub n = 1 it'll obviously be <= -1
step 2, assume true for n = k
k(k+1)(k+2)(k+3) >= -1
and prove true for n = k+1
ie, LHS = (k+1)(K+2)(k+3)(k+4)
>= (-1/k)(k+4) [from assumption]
and so even if you minus something, the LHS is still greater either way
so..
>= -1

I think that's right..

EDIT: i think the -1/k is a bit controversial so better wait for another reply

It's for all real numbers x, not just x >= 0

 

[conics2008]

shoudn't there be atleast more things with that question.

 

[lyounamu]

shoudn't there be atleast more things with that question.

n(n+1)(n+2)(n+3) >= -1
(n+1)(n+2)(n+3) >= -1/n

I got up to where we say:

L.H.S = (n+1)(n+2)(n+3)(n+4)
= n(n+1)(n+2)(n+3) + 4(n+1)(n+2)(n+3)
But what can I do from here?

Damn, get rid of all those rubbish here we go:

x(x+1)(x+2)(x+3) = (x^2+x)(x^2+5x+6) = x^4 + 5x^3 + 6x^2 + x^3 + 5x^2 + 6x = x^4 + 6x^3 + 11x^2 + 6x
f'(x) = 4x^3 + 18x^2 + 22x + 6
= 2(2x^3+9x^2+11x+3)
Using approximation method, f'(x) = 0 when x = -0.38

Minimum is when x = -0.38 so y = -0.38(-0.38 +1)(-0.38+2)(-0.38+3) = -0.999999806...>=-1

 

[conics2008]

ly ... inequalities you can just improve it by induction...

there are properties which you prove at the begining.. to me it looks like an incomplete question.

 

[Just.Snaz]

It's for all real numbers x, not just x >= 0

true.. but then i have no clue how to do it... start with x^2 >= 0 ? i couldn't use that effectively though..

 

[vds700]

you could solve it by sketching y = x(x + 1)(x + 2)(x + 3) and showing that it lies above y = -1 couldn't you?

EDIT: Just saw Namu's post. Good work mate, though there are 2 stationary points (i plotted it using winplot hehe)

 

[lyounamu]

you could solve it by sketching y = x(x + 1)(x + 2)(x + 3) and showing that it lies above y = -1 couldn't you?

Yeah, that's what I did. But you cannot really "show" it by drawing it.

I think you have to use differentiation and find the minimum point and show that the minimum point is >= -1...

 

[vds700]

Yeah, that's what I did. But you cannot really "show" it by drawing it.

I think you have to use differentiation and find the minimum point and show that the minimum point is >= -1...

still i reckon its best to find both minimum SP's and draw the graph, it clearly shows the result

 

[lyounamu]

still i reckon its best to find both minimum SP's and draw the graph, it clearly shows the result

By the way, if you have derivative like this:

f'(x) = 4x^3 + 18x^2 + 22x + 6,

how would you come about finding the x when f'(x) = 0 ?

I used the approximation method and that seemed to have got me a very very close answer.

 

[ctpengage]

ly ... inequalities you can just improve it by induction...

there are properties which you prove at the begining.. to me it looks like an incomplete question.

nup its complete. can anyone solve it without using calculus or induction?

 

[lyounamu]

nup its complete. can anyone solve it without using calculus or induction?

As far as I am aware, using induction and calculus are the best way to get to the solution.

 

[midifile]

Okay.. so you know that ((x + 1/x) + 3)² >= 0 (as anything squared >= 0)
Therefore by expanding:
(x + 1/x)² + 6(x + 1/x) + 9 >= 0
By expanding again:
x² + 2 + 1/x² + 6x + 6/x + 9 >= 0
x² + 1/x² + 6x + 6/x +11 >= 0

But x² is also >= 0
So x²(x² + 6x + 11 + 6/x + 1/x²) >= 0
Therefore x⁴ + 6x³ + 11x² + 6x + 1 >= 0
x⁴ + 6x³ + 11x² + 6x >= -1

The you can factorise the lhs to get x(x+1)(x+2)(x+3)
So x(x+1)(x+2)(x+3) >= -1

 

[lyounamu]

Okay.. so you know that ((x + 1/x) + 3)² >= 0 (as anything squared >= 0)
Therefore by expanding:
(x + 1/x)² + 6(x + 1/x) + 9 >= 0
By expanding again:
x² + 2 + 1/x² + 6x + 6/x + 9 >= 0
x² + 1/x² + 6x + 6/x +11 >= 0

But x² is also >= 0
So x²(x² + 6x + 11 + 6/x + 1/x²) >= 0
Therefore x⁴ + 6x³ + 11x² + 6x + 1 >= 0
x⁴ + 6x³ + 11x² + 6x >= -1

The you can factorise the lhs to get x(x+1)(x+2)(x+3)
So x(x+1)(x+2)(x+3) >= -1

Damn, how did you get all that sorted out?

 

[midifile]

Damn, how did you get all that sorted out?

Well originally I went backwards - I expanded x(x+1)(x+2)(x+2) and factorised till I got (x + 1/x + 3)², and then that is where I started from and worked forwards.

My 4unit teacher says with inequalities like this you will not get the marks if you assume the inequality is true, so you have to find something that is true, and then manipulate it to get the inequality.

But your method was good too, especially considering you havent done 4u inequalities yet

 

[lyounamu]

Well originally I went backwards - I expanded x(x+1)(x+2)(x+2) and factorised till I got (x + 1/x + 3)², and then that is where I started from and worked forwards.

My 4unit teacher says with inequalities like this you will not get the marks if you assume the inequality is true, so you have to find something that is true, and then manipulate it to get the inequality.

But your method was good too, especially considering you havent done 4u inequalities yet

So if I could actually get the right value of x, (not the estimation) would my answer be correct?

 

[midifile]

Yeah im pretty sure you would. We've done some inequalities in class where we have just had to prove that the minimum (or maximum) turning point is above (or below) a value.

The only problem is if you are unable to find the actual root, you may spend ages going around in circles on a question that is only worth 3 or 4 marks.

 

[lyounamu]

Yeah im pretty sure you would. We've done some inequalities in class where we have just had to prove that the minimum (or maximum) turning point is above (or below) a value.

The only problem is if you are unable to find the actual root, you may spend ages going around in circles on a question that is only worth 3 or 4 marks.

Yeah, true.

 

[lolokay]

let x = (v-3)/2

(v-3)/2(v-1)/2(v+1)/2(v+3)/2 >= -1
(v+3)(v-3)(v+1)(v-1) >= -16
v⁴ - 10v² + 9 >= -16
v⁴ - 10v² + 25 >= 0
(v² - 5)² >= 0
:. true