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bernoulli's inequality question

show using the inequality that

∀ x≥1\forall\ x\geq 1

21x≤1+1x2^{\frac{1}{x}} \leq 1+\dfrac{1}{x}

i know bernoullis inequality is


(1+x)r≥1+rx(1+x)^r \geq 1+rx

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

∀ x≥1\forall\ x\geq 1

21x≤1+1x2^{\frac{1}{x}} \leq 1+\dfrac{1}{x}

i know bernoullis inequality is


(1+x)r≥1+rx(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≤1+ry(1+y)^r \leq 1+ry for 0≤r≤10 \leq r \leq 1 and y≥−1y \geq -1

Now work with this.
Original post by RDKGames
Gonna replace the variables in the ineq. and say that (1+y)r≤1+ry(1+y)^r \leq 1+ry for 0≤r≤10 \leq r \leq 1 and y≥−1y \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−2x 1+x^{-2} x but i'm not sure about the right hand side
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−2x 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≥−1y \geq -1 and r=f(x)∈[0,1]r=f(x) \in [0,1] (for x≥1x \geq 1) in order to get 21/x2^{1/x} on the LHS. Then the RHS just falls out.
Original post by RDKGames
Rather than doing whatever you tried to do here, I'd recommend picking appropriate y≥−1y \geq -1 and r=f(x)∈[0,1]r=f(x) \in [0,1] (for x≥1x \geq 1) in order to get 21/x2^{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?
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=1y=1 is fine, but is r=1xr=\frac{1}{x} what you're looking for to substitute. Agree?
Original post by RDKGames
y=1y=1 is fine, but is r=1xr=\frac{1}{x} what you're looking for to substitute. Agree?


uh ye, so r= 0.x?
Original post by will'o'wisp2
uh ye, so r= 0.x?


Huh?
Original post by RDKGames
Huh?


what am i doing with r?
Original post by will'o'wisp2
what am i doing with r?


Replacing it by 1x\frac{1}{x} in the inequality, obviously.
Original post by RDKGames
Replacing it by 1x\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?
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,yr,y such that, for x≥1x \geq 1, r=1x∈[0,1]r=\frac{1}{x} \in [0,1] and y=1≥−1y=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.
(edited 6 years ago)
Original post by RDKGames
You've used a proven result (Bernoulli's ineq.) to show the proposed result by picking appropriate r,yr,y such that, for x≥1x \geq 1, r=1x∈[0,1]r=\frac{1}{x} \in [0,1] and y=1≥−1y=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
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≤1+xy(1+y)^x \leq 1+xy for x∈[0,1]x \in [0,1] and y≥−1y \geq -1.

Pick y=1y=1, then 2x≤1+x2^x \leq 1+x which is true by the above theorem.

Then doing x↦1xx \mapsto \frac{1}{x} means 21/x≤1+1x\displaystyle 2^{1/x} \leq 1+\frac{1}{x} if 1x∈[0,1]\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 xx from solving 0≤1x≤10 \leq \frac{1}{x} \leq 1

Perhaps you're more comfortable with this approach.
(edited 6 years ago)
Original post by RDKGames
You may also prove it by saying that:

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

Pick y=1y=1, then 2x≤1+x2^x \leq 1+x which is true by the above theorem.

Then doing x↦1xx \mapsto \frac{1}{x} means 21/x≤1+1x\displaystyle 2^{1/x} \leq 1+\frac{1}{x} if 1x∈[0,1]\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 xx from solving 0≤1x≤10 \leq \frac{1}{x} \leq 1

Perhaps you're more comfortable with this approach.


https://cdn.discordapp.com/attachments/322819040059850757/379380070751535105/unknown.png

i don't understand where the 1+ x multiplied by 1/x comes from
Original post by will'o'wisp2
https://cdn.discordapp.com/attachments/322819040059850757/379380070751535105/unknown.png

i don't understand where the 1+ x multiplied by 1/x comes from


(1+y)r≥1+ry(1+y)^r \geq 1+ry is Bernoulli's so let r=xr=x and y=1xy=\frac{1}{x} which gives the result.
Original post by RDKGames
(1+y)r≥1+ry(1+y)^r \geq 1+ry is Bernoulli's so let r=xr=x and y=1xy=\frac{1}{x} which gives the result.


Oh i see thanks

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