Is x=0 a solution to the inequality above?

[Rectifier]

Homework Statement

$$x+\frac{16}{\sqrt{x}} \geq 12$$

How do I show that only x>0 satisfies the inequality above.

Homework Equations

The Attempt at a Solution

I have not made a lot of progress here. I tried the following:
$$x+\frac{16}{\sqrt{x}} - 12 \geq 0$$

I tried to multiply with $$ \sqrt{x} $$ but i am not sure wether that is allowed or not.

I am stuck here :,(

 

[ehild]

Homework Statement

$$x+\frac{16}{\sqrt{x}} \geq 12$$

How do I show that only x>0 satisfies the inequality above.

Is the expression on the left side defined for x≤0?

 

[Rectifier]

Is the expression on the left side defined for x≤0?

I don't think so. There were no additional information in the problem so i suppose that we should use real numbers. And thus for $\sqrt{x}$ to be > 0 the x should be more that 0. Am I right? Was that problem that easy? :O

 

[ehild]

I don't think so. There were no additional information in the problem so i suppose that we should use real numbers. And thus for $\sqrt{x}$ to be > 0 the x should be 0 or more. Am I right? Was that problem that easy? :O

For $\sqrt{x}$ to be > 0, x should be definitely greater then zero. You can not divide by zero!
That is not the solution yet, only a condition x must satisfy. You have to show for what positive x values the inequality is valid. You can introduce a new variable z=√x, arrange the inequality in the form f(z)≥0 and try to factorize f(z), by finding the roots. (They are simple integers )

 

[Rectifier]

For $\sqrt{x}$ to be > 0, x should be definitely greater then zero. You can not divide by zero!
That is not the solution yet, only a condition x must satisfy. You have to show for what positive x values the inequality is valid. You can introduce a new variable z=√x, arrange the inequality in the form f(z)≥0 and try to factorize f(z), by finding the roots. (They are simple integers )

Thank you for your help.

I have set $z=\sqrt{x}$

$$z^2+\frac{16}{z} \geq 12 $$

$$z^2+\frac{16}{z} -12 \geq 0 $$

And I get stuck :,(

 

[Ray Vickson]

Homework Statement

$$x+\frac{16}{\sqrt{x}} \geq 12$$

How do I show that only x>0 satisfies the inequality above.

Homework Equations

The Attempt at a Solution

I have not made a lot of progress here. I tried the following:
$$x+\frac{16}{\sqrt{x}} - 12 \geq 0$$

I tried to multiply with $$ \sqrt{x} $$ but i am not sure wether that is allowed or not.

I am stuck here :,(

Minimize the function $f(x) = x + 16/\sqrt{x}$ in the region $x > 0$. What do you get?

 

[Ray Vickson]

I don't think so. There were no additional information in the problem so i suppose that we should use real numbers. And thus for $\sqrt{x}$ to be > 0 the x should be more that 0. Am I right? Was that problem that easy? :O

NO, not necessarily: what is stopping $x + 16/\sqrt{x}$ from being < 12 for some $x> 0$? You need to show that cannot happen!

 

[Rectifier]

Minimize the function $f(x) = x + 16/\sqrt{x}$ in the region $x > 0$. What do you get?

Alright, then:
$$ f(x)=x+ \frac{16}{\sqrt{x}} \\ f'(x)=1- \frac{8}{x^{\frac{3}{2}}} \\ 0=1- \frac{8}{x^{3}{2}} \\ x=4 $$

I am afraid that I don't know how that x=4 is going to make my life easier :D

 

[Ray Vickson]

Alright, then:
$$ f(x)=x+ \frac{16}{\sqrt{x}} \\ f'(x)=1- \frac{8}{x^{\frac{3}{2}}} \\ 0=1- \frac{8}{x^{3}{2}} \\ x=4 $$

I am afraid that I don't know how that x=4 is going to make my life easier :D

Yes, but knowing that $f(4)$ is the smallest possible of $f(x)$ in the region $x > 0$ is valuable information, especially since evaluating $f(4)$ is not very hard.

 

[Rectifier]

Yes, but knowing that $f(4)$ is the smallest possible of $f(x)$ in the region $x > 0$ is valuable information, especially since evaluating $f(4)$ is not very hard.

How can I use that value to solve the problem? :D I am sorry if I am being slow on this one.

 

[Rectifier]

How can I use that value to solve the problem? :D I am sorry if I am being slow on this one.

The interesting thing here is that i get 12 if I try to calculate f(4). This means that 12 is the lowest value the function can have for x>0!

 

[Ray Vickson]

The interesting thing here is that i get 12 if I try to calculate f(4). This means that 12 is the lowest value the function can have for x>0!

Exactly: so you have proved the statement they asked you to prove!

 

[Rectifier]

Exactly: so you have proved the statement they asked you to prove!

Yey! :D
Oh gosh I am so appy now. I have been sitting with this one for about a day now. Thank you Ray and thank you ehild! :D

 

[SammyS]

Homework Statement

$$x+\frac{16}{\sqrt{x}} \geq 12$$

How do I show that only x>0 satisfies the inequality above ?

I don't see that this original question was answered.

ehild did try to lead you to the answer.

Suppose that x < 0, i.e. x is negative.

What is $\ \sqrt{-4\,}\$ ?

 

[Rectifier]

I don't see that this original question was answered.

ehild did try to lead you to the answer.

Suppose that x < 0, i.e. x is negative.

What is $\ \sqrt{-4\,}\$ ?

2i

 

[SammyS]

2i

IGNORE

Of course that's not a real number.

Oh SNAP !

I see you did answer this in Post #3 .

 

[Ray Vickson]

I don't see that this original question was answered.

ehild did try to lead you to the answer.

Suppose that x < 0, i.e. x is negative.

What is $\ \sqrt{-4\,}\$ ?

I think the question was somewhat weirdly stated. Certainly, the inequality makes no sense at all if $x < 0$, but for $x > 0$ there is still a non-trivial issue to prove, viz., that $f(x) \geq 12\; \forall \, x > 0$.

 

[ehild]

With the notation z=√x, the original inequality can be written as z³-12z+16 ≥ 0. In order to factorize the left side, we try to find one root. As the coefficients are all integer, it is possible that integer root exist, one among the dividers of 16. 1 is not a root, but 2 is, and -4 also. So the inequality can be written as (z-2)²(z+4) ≥ 0
Knowing that z>0, is there any value of z so that the product (z-2)²(z+4) becomes negative?

 

[vela]

The interesting thing here is that i get 12 if I try to calculate f(4). This means that 12 is the lowest value the function can have for x>0!

You've only shown x=4 is a critical point so far. For completeness, you should show that f(4) is a minimum.

 

[Rectifier]

With the notation z=√x, the original inequality can be written as z³-12z+16 ≥ 0. In order to factorize the left side, we try to find one root. As the coefficients are all integer, it is possible that integer root exist, one among the dividers of 16. 1 is not a root, but 2 is, and -4 also. So the inequality can be written as (z-2)²(z+4) ≥ 0
Knowing that z>0, is there any value of z so that the product (z-2)²(z+4) becomes negative?

why is it ok to multiply with z when going from:
$z^2+ \frac{16}{z} - 12 \geq 0$
to
$z^3+16-12z ≥ 0$ ?

Thought:
We know that z>0 right now. So I guess that it is allowed. But what if z could be z<0 or perhaps even z=0 (thats not the case this time) but what if? :)

I am sorry if I am going slightly off-topic.

 

[SammyS]

why is it ok to multiply with z when going from:
$z^2+ \frac{16}{z} - 12 \geq 0$
to
$z^3+16-12z ≥ 0$ ?

Thought:
We know that z>0 right now. So I guess that it is allowed. But what if z could be z<0 or perhaps even z=0 (that's not the case this time) but what if? :)

I am sorry if I am going slightly off-topic.

There are a number of methods to handle this.

Other methods can be quicker, but the most general method is to consider the possible cases:

If z > 0 : then ...

If z < 0 : then ...

(Of course, z ≠ 0 for examples like this.)

 

[ehild]

why is it ok to multiply with z when going from:
$z^2+ \frac{16}{z} - 12 \geq 0$
to
$z^3+16-12z ≥ 0$ ?

Thought:
We know that z>0 right now. So I guess that it is allowed. But what if z could be z<0 or perhaps even z=0 (thats not the case this time) but what if? :)

I am sorry if I am going slightly off-topic.

Z can not be zero as it is in the denominator.
If z may be negative you have to discuss both cases. In case it is negative and you multiply with it, the inequality will change: $z^3+16-12z ≤ 0$ , that is, (z-2)²(z+4)≤0, that can happen if ? So $z^2+ \frac{16}{z} - 12 \geq 0$ is satisfied by z from two intervals.