Cool solution.

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Problem: Prove that\[\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\]for distinct reals \(a,b\).
How many seconds can I have to think about it before you give the solution? :p
Try it out. Just meant to share this problem.

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Other answers:

\[(\frac{1+a}{1+b})^{\frac{1}{a-b}} \\ =(\frac{1+b+a-b}{1+b})^\frac{1}{a-b} \\ =(1+\frac{a-b}{1+b})^{\frac{1}{a-b}}\] \[=(1+\frac{1}{\frac{1+b}{a-b}})^\frac{1}{a-b} \\ =(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^\frac{1}{a-b}\] \[\lim_{a-b \rightarrow \infty}(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^{\frac{1}{a-b}}=\lim_{a-b \rightarrow \infty} e^{\frac{1}{1+b}} \\ ...\] if a-b goes to infinity I don't think we can say anything about the limit of b from that so that limit should be just exp(1/(1+b)) ..I don't know if this is the right route For some reason this is the first thing that popped in my head
Ganeshie is here - he'll easily crack this one. :|
I did something wrong
I should have said 1/(a-b) goes to infinity
which would mean a-b goes to 0
`ganeshie8 is typing a reply...` Well, here we go.
$$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\\\frac1{a-b}\log\left(\frac{1+a}{1+b}\right)<1\\\frac1{a-b}\left(\log(1+a)-\log(1+b)\right)<1\\\log(1+a)-\log(1+b)<(1+a)-(1+b)$$... is just a restatement of the fact that \(\log\) is concave by Jensen's inequality
all those steps are reversible, so the proof just follows from the fact that \(\log\) is concave via Jensen's inequality
I don't think my way is good enough it only shows for a close to b we have exp(1/(1+b)) but I think this right here is
\[e^{x-y} \gt 1+(x-y)+(x-y)^2/2 \gt 1+x-y = 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]
haha that one is great
but hmm @ganeshie8 both you and i have only proven it for \(a-b>0\) technically
Ah right, I see..
unless you can rigorously show that the even terms outweigh the odd ones in the expansion \(e^x=1+x+\frac12 x^2+\dots\)
https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder
looks the taylor series one also requires \(y\gt 0\) \[e^{x-y} \gt 1+(x-y)\gt 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}\]
@parthkohli share your answer
@ganeshie8 That's exactly my answer. I think I got the incomplete question.
Hey, your next project is to implement LaTeX in the drawing tool. That looks good.
looks the inequality is true only when \(a,b\) have same signs |dw:1437673377179:dw|
So yes, I did get an incomplete question. You can't expect much from Facebook math groups.
Haha okay but my/your solution is incomplete too even after restricting to first quadrant because it requires \(a-b\gt 0\) as pointed out by oldrin earlier

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