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    show using the inequality that

    \forall\ x\geq 1

    2^{\frac{1}{x}} \leq 1+\dfrac{1}{x}

    i know bernoullis inequality is


    (1+x)^r \geq 1+rx

    but i'm not seeing how i can prove the above one using this one here
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    (Original post by will'o'wisp2)
    show using the inequality that

    \forall\ x\geq 1

    2^{\frac{1}{x}} \leq 1+\dfrac{1}{x}

    i know bernoullis inequality is


    (1+x)^r \geq 1+rx

    but i'm not seeing how i can prove the above one using this one here
    Gonna replace the variables in the ineq. and say that (1+y)^r \leq 1+ry for 0 \leq r \leq 1 and y \geq -1

    Now work with this.
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    (Original post by RDKGames)
    Gonna replace the variables in the ineq. and say that (1+y)^r \leq 1+ry for 0 \leq r \leq 1 and y \geq -1

    Now work with this.
    so the left hand side i can make 1+1/x in the form 1+rx by saying that  1+x^{-2} x but i'm not sure about the right hand side
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    (Original post by will'o'wisp2)
    so the left hand side i can make 1+1/x in the form 1+rx by saying that  1+x^{-2} x but i'm not sure about the right hand side
    Rather than doing whatever you tried to do here, I'd recommend picking appropriate y \geq -1 and r=f(x) \in [0,1] (for x \geq 1) in order to get 2^{1/x} on the LHS. Then the RHS just falls out.
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    (Original post by RDKGames)
    Rather than doing whatever you tried to do here, I'd recommend picking appropriate y \geq -1 and r=f(x) \in [0,1] (for x \geq 1) in order to get 2^{1/x} on the LHS. Then the RHS just falls out.
    so i picked y=1 which gives me 2 but i can't think of an r within 0 and 1 inclusive so that r = 1/x i just can't find one


    so i thought about 0.5 which is like to the power of 1/2 which is great but how do i do 1/x?
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    (Original post by will'o'wisp2)
    so i picked y=1 which gives me 2 but i can't think of an r within 0 and 1 inclusive so that r = 1/x i just can't find one


    so i thought about 0.5 which is like to the power of 1/2 which is great but how do i do 1/x?
    y=1 is fine, but is r=\frac{1}{x} what you're looking for to substitute. Agree?
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    (Original post by RDKGames)
    y=1 is fine, but is r=\frac{1}{x} what you're looking for to substitute. Agree?
    uh ye, so r= 0.x?
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    (Original post by will'o'wisp2)
    uh ye, so r= 0.x?
    Huh?
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    (Original post by RDKGames)
    Huh?
    what am i doing with r?
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    (Original post by will'o'wisp2)
    what am i doing with r?
    Replacing it by \frac{1}{x} in the inequality, obviously.
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    (Original post by RDKGames)
    Replacing it by \frac{1}{x} in the inequality, obviously.
    uh so i just get out what i question originally says, so is that the proof? finding values of y and r such that you can make the inequality in the original question?
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    (Original post by will'o'wisp2)
    uh so i just get out what i question originally says, so is that the proof? finding values of y and r such that you can make the inequality in the original question?
    You've used a proven result (Bernoulli's ineq.) to show the proposed result by picking appropriate r,y such that, for x \geq 1, r=\frac{1}{x} \in [0,1] and y=1\geq -1 thus satisfying Bernoulli's requirements for it to hold true, so your result must hold true. Since your result is what you're asked to prove, then well... you've proven it.

    There might be a different approach but this one works out quite nicely.
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    (Original post by RDKGames)
    You've used a proven result (Bernoulli's ineq.) to show the proposed result by picking appropriate r,y such that, for x \geq 1, r=\frac{1}{x} \in [0,1] and y=1\geq -1 thus satisfying Bernoulli's requirements for it to hold true, so your result must hold true. Since your result is what you're asked to prove, then well... you've proven it.

    There might be a different approach but this one works out quite nicely.
    ah ok coolio, thanks man
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    (Original post by will'o'wisp2)
    ah ok coolio, thanks man
    You may also prove it by saying that:

    Bernoulli's inequality is (1+y)^x \leq 1+xy for x \in [0,1] and y \geq -1.

    Pick y=1, then 2^x \leq 1+x which is true by the above theorem.

    Then doing x \mapsto \frac{1}{x} means \displaystyle 2^{1/x} \leq 1+\frac{1}{x} if \frac{1}{x} \in [0,1] which leads us to the condition that the statement is true only if ....?

    And just finish it off by stating the condition on x from solving 0 \leq \frac{1}{x} \leq 1

    Perhaps you're more comfortable with this approach.
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    (Original post by RDKGames)
    You may also prove it by saying that:

    Bernoulli's inequality is (1+y)^x \leq 1+xy for x \in [0,1] and y \geq -1.

    Pick y=1, then 2^x \leq 1+x which is true by the above theorem.

    Then doing x \mapsto \frac{1}{x} means \displaystyle 2^{1/x} \leq 1+\frac{1}{x} if \frac{1}{x} \in [0,1] which leads us to the condition that the statement is true only if ....?

    And just finish it off by stating the condition on x from solving 0 \leq \frac{1}{x} \leq 1

    Perhaps you're more comfortable with this approach.
    https://cdn.discordapp.com/attachmen...05/unknown.png

    i don't understand where the 1+ x multiplied by 1/x comes from
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    (Original post by will'o'wisp2)
    https://cdn.discordapp.com/attachmen...05/unknown.png

    i don't understand where the 1+ x multiplied by 1/x comes from
    (1+y)^r \geq 1+ry is Bernoulli's so let r=x and y=\frac{1}{x} which gives the result.
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    (Original post by RDKGames)
    (1+y)^r \geq 1+ry is Bernoulli's so let r=x and y=\frac{1}{x} which gives the result.
    Oh i see thanks
 
 
 
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