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Inequality satisfaction

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  1. Jun 7, 2015 #1

    Rectifier

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    1. The problem statement, all variables and given/known data
    $$x+\frac{16}{\sqrt{x}} \geq 12$$

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

    2. Relevant equations


    3. 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 :,(
     
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  3. Jun 7, 2015 #2

    ehild

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    Is the expression on the left side defined for x≤0?
     
  4. Jun 7, 2015 #3

    Rectifier

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    I dont 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
     
    Last edited: Jun 7, 2015
  5. Jun 7, 2015 #4

    ehild

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    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 :smile: )
     
  6. Jun 7, 2015 #5

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    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 :,(
     
  7. Jun 7, 2015 #6

    Ray Vickson

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    Minimize the function ##f(x) = x + 16/\sqrt{x}## in the region ##x > 0##. What do you get?
     
  8. Jun 7, 2015 #7

    Ray Vickson

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    NO, not necessarily: what is stopping ##x + 16/\sqrt{x}## from being < 12 for some ##x> 0##? You need to show that cannot happen!
     
  9. Jun 7, 2015 #8

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    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
     
  10. Jun 7, 2015 #9

    Ray Vickson

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    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.
     
  11. Jun 7, 2015 #10

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    How can I use that value to solve the problem? :D I am sorry if I am being slow on this one.
     
  12. Jun 7, 2015 #11

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    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!
     
  13. Jun 7, 2015 #12

    Ray Vickson

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    Exactly: so you have proved the statement they asked you to prove!
     
  14. Jun 7, 2015 #13

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    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
     
  15. Jun 7, 2015 #14

    SammyS

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    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\,}\ ## ?
     
  16. Jun 7, 2015 #15

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    2i
     
  17. Jun 7, 2015 #16

    SammyS

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    IGNORE

    Of course that's not a real number.

    Oh SNAP !

    I see you did answer this in Post #3 .
     
    Last edited: Jun 7, 2015
  18. Jun 7, 2015 #17

    Ray Vickson

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    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##.
     
  19. Jun 7, 2015 #18

    ehild

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    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?
     
  20. Jun 7, 2015 #19

    vela

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    You've only shown x=4 is a critical point so far. For completeness, you should show that f(4) is a minimum.
     
  21. Jun 8, 2015 #20

    Rectifier

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    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.
     
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