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     \displaystyle u_1 = 5.2, \text{and } u_{n+1} = \frac{6u_n+25}{u_n+6} .

    Prove by induction that  u_n>5 .
    How do you deal with these inequality types? It seems so trivial but how to give a sound proper proof. It seems that you rearrange it to
     u_{n+1} = \displaystyle 6-\frac{11}{u_n+6} .
    Then it seems trivial.
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    (Original post by Ano123)
     \displaystyle u_1 = 5.2, \text{and } u_{n+1} = \frac{6u_n+25}{u_n+6} .

    Prove by induction that  u_n>5 .
    As a general rule:

    1. Show it works for case n=1 or another easy to evaluate one.

    2. Assume it works for n=k.

    3. Show that it must then work for n=k+1, normally involves relating to the n=k case.

    4. Hence holds by induction
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    (Original post by samb1234)
    As a general rule:

    1. Show it works for case n=1 or another easy to evaluate one.

    2. Assume it works for n=k.

    3. Show that it must then work for n=k+1, normally involves relating to the n=k case.

    4. Hence holds by induction
    What would you make the inductive step?
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    (Original post by Ano123)
    What would you make the inductive step?
    Showing that the statement holds for n=k+1 given that it holds for n=k.
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    (Original post by Ano123)
    What would you make the inductive step?
    Assume that Uk>5, then using your rearranged form for Un+1 can easily show that if Uk is greater than 5 than so is Uk+1
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    (Original post by samb1234)
    Assume that Uk>5, then using your rearranged form for Un+1 can easily show that if Uk is greater than 5 than so is Uk+1
    That's what I've done and demonstrated in the OP, but how can I formally say it. It doesn't seem right saying, it is clearly greater than 5 as long as  u_n > 5 .
    But in this case do you think it's so obvious that it is fine to just do that?
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    (Original post by Ano123)
    That's what I've done and demonstrated in the OP, but how can I formally say it. It doesn't seem right saying, it is clearly greater than 5 as long as  u_n > 5 .
    But in this case do you think it's so obvious that it is fine to just do that?
    Maybe make a statement that 11/Uk+6 is less than 1 if Uk is greater than 5, and therefore 6 -11/Uk+6 is always greater than 5 if holds for Uk
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    (Original post by samb1234)
    Maybe make a statement that 11/Uk+6 is less than 1 if Uk is greater than 5, and therefore 6 -11/Uk+6 is always greater than 5 if holds for Uk
    Thank you for the help.
 
 
 
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