Proving inequality (-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2 < 0)

[BJU]

I have an expression that I need to prove is negative for 0<x<1.

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2 < 0

I can plot the left hand side and see that it is indeed true, but I do not know how to prove it. I tried to use polynomial division, but to no avail.

If anybody has a suggestion, I would be very grateful.

Cheers,
BJ

 

[Deleted member 4993]

I have a term that I need to prove is negative for 0<x<1.

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2 < 0

I can plot the left hand side and see that it is indeed true, but I do not know how to prove it. I tried to use polynomial division, but to no avail.

If anybody has a suggestion, I would be very grateful.

Cheers,
BJ

I have not tried it - but if I had to do the problem, I would substitute:

u = √(4-3x).... show us what do you get.

 

[BJU]

I have not tried it - but if I had to do the problem, I would substitute:

u = √(4-3x).... show us what do you get.

Ah, substitution, I should have thought of that. I'll give it a try and let you know.

Many thanks!

 

[BJU]

Ok, substitution did not really solve the issue, but it gave me the idea to isolate the square root on one side (that should have been an obvious approach in the first place), square both sides and go from there. Eventually, I ended up with

(1-x)(1-x)(667 - 569 x + 33 x^2 - 3 x^3) > 0

which is clearly true for 0<x<1.

 

[Deleted member 4993]

Ok, substitution did not really solve the issue, but it gave me the idea to isolate the square root on one side (that should have been an obvious approach in the first place), square both sides and go from there. Eventually, I ended up with

(1-x)(1-x)(667 - 569 x + 33 x^2 - 3 x^3) > 0

which is clearly true for 0<x<1.

I think, you have proved that the square of an expression is "positive" - but that does not prove the expression itself is positive in the given domain.

 

[BJU]

I think, you have proved that the square of an expression is "positive" - but that does not prove the expression itself is positive in the given domain.

I realize you can't just always square both sides, but if both sides are positive, is it not fair game to do it?

If I solve for Sqrt[4-3x] I get

Sqrt[4-3x] < (55-50x+3x^2) / (16-8x)

and both sides are positive for 0<x<1, so I figured I'm ok.

 

[BJU]

Is that the same as:
8 * x * Sqrt[4 - 3 * x] ?

Yes, that's right. Sorry for the unusual notation.

 

[Deleted member 4993]

I have an expression that I need to prove is negative for 0<x<1.

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2 < 0

I can plot the left hand side and see that it is indeed true, but I do not know how to prove it. I tried to use polynomial division, but to no avail.

If anybody has a suggestion, I would be very grateful.

Cheers,
BJ

Substituting u = √(4-3x), we get:

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2
= 19 + 16u - 42u^2 + 8u^3 - u^4

= -(u - 1) (u^3 - 7 u^2 + 35 u + 19)

= -(u-1)[(u-2)^3 + (-u^2 + 23u + 27)]


= -(u-1)[(u-2)^3 + (-u^2 + 23u + 27)]


= -(u-1)[(u-2)^3 - (u - 2)^2 + 23(u -2) + 73]..............where 1 < u < 2

The second factor is >0 hence......

 

[mmm4444bot]

(1-x)(1-x)(667 - 569 x + 33 x^2 - 3 x^3) > 0

which is clearly true for 0 < x < 1

You solved:

(1 - x)^2 > 0

to show that x must be less than 1, yes?

How did you solve:

667 - 569x + 33x^2 - 3x^3 > 0 ?

That is, you have posted that the product (1-x)^2*(667-569x+33x^2-3x^3) is "clearly" positive, when 0<x<1, but you have not demonstrated how you proved this statement.

When we're asked to prove something, we need to explain such conclusions. :cool:

 

[mmm4444bot]

Sorry for the unusual notation.

No need to apologize. (I'm not sure why Denis was concerned.)

The Commutative Property of Multiplication tells us that we may multiply objects in any order. Each of the following expressions is equivalent.

-8*sqrt(4-3x)*x

-8*x*sqrt(4-3*x)

sqrt(4-3x)*(-8)*x

sqrt(4-3x)*x*(-8)

x*sqrt(4-3x)*(-8)

x*(-8)*sqrt(4-3x)

:cool:

 

[mmm4444bot]

Re-arranging slightly (and setting = 0):

3*x^2 + 8*x*sqrt(4 - 3*x) - 16*sqrt(4 - 3*x) + 55 = 0

Then: x = ~-24.758 or x = ~-.809

This shows that the given expression is greater than zero (within the given domain), not less. Your rearranging must be slightly off.

Compliments of Wolfram

Is this allowed, in BJU's class?

 

[mmm4444bot]

… if both sides are positive, is it not fair game to [square each side of an inequality]?

No, I think that's okay. However, keep in mind that squaring relationships can introduce extraneous solutions. In your exercise, the extra solutions created are not an issue because they occur outside of the given domain.

 

[mmm4444bot]

I used Wolfram simply as a calculator.

Excusez-moi; I was thinking that wolfram solved your equation for you.

By the way, I found your mistake. Your equation is missing the -50x term; solution is now x = 1.

What's the next step, in your proof? :cool:

 

[BJU]

Substituting u = √(4-3x), we get:

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2
= 19 + 16u - 42u^2 + 8u^3 - u^4
...

Ah, that's really neat, I did not consider that. I'll keep that technique in mind. I have lots of similar problems that hopefully I can solve with this.

 

[BJU]

Is this allowed, in BJU's class?

This is not actually for a class, but a paper (I'm an experimental economist and my math is quite rusty in places).

I already had the numerical solution (I use Mathematica ), but that always looks a bit bad, so I was looking for a proper proof.


Edit: And completely off-topic:
"English is the most ambiguous language in the world." ~ Yours Truly, 1969

I think Chinese is the most ambiguous language. Translating into German (probably the least ambiguous language in the world), much less meaning is lost when it's English than when it's Chinese.

 

[BJU]

How did you solve:

667 - 569x + 33x^2 - 3x^3 > 0 ?

That is, you have posted that the product (1-x)^2*(667-569x+33x^2-3x^3) is "clearly" positive, when 0<x<1, but you have not demonstrated how you proved this statement.

When we're asked to prove something, we need to explain such conclusions. :cool:

I figured since the - 569x - 3x^3 has a minimum of -572, it cannot compensate for the 667 and that this is obvious enough to stop there. Do you think it is better / more appropriate to further reduce it to

(1-x)((1-x)(27-3x)+512) + 128

?

 

[Deleted member 4993]

Guilty as charged. How long in the corner?

50 minutes.....

 

[JeffM]

I have an expression that I need to prove is negative for 0<x<1.

-55 + 16 Sqrt[4 - 3 x] + 50 x - 8 Sqrt[4 - 3 x] x - 3 x^2 < 0

I can plot the left hand side and see that it is indeed true, but I do not know how to prove it. I tried to use polynomial division, but to no avail.

If anybody has a suggestion, I would be very grateful.

Cheers,
BJ

I do not have time to work this out this morning. Take the derivative of the function. Show that it is positive in the interval (0, 1). Calculate the function at 0 and 1. I suspect that will do it.

 

[BJU]

I do not have time to work this out this morning. Take the derivative of the function. Show that it is positive in the interval (0, 1). Calculate the function at 0 and 1. I suspect that will do it.

Thanks for the suggestions.

I had actually tried that approach, but the derivative just creates a variation of the original problem, so it does not work in this particular case.

 

[mmm4444bot]

I had actually tried that approach, but the derivative just creates a variation of the original problem, so it does not work in this particular case.

Yes, I see what you mean.

Using the derivative, I ended up needing to solve:

sqrt(4 - 3x) > 2(9x - 14)/(3x - 25)

I have a hunch that the derivative approach may be workable, but I need to consider this some more.

For now, I like Subhotosh's method because I can explain by analysis why the following is true (for u between 1 and 2).

(u-2)^3 - (u-2)^2 + 96 > 0

:cool: