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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.

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.

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..

https://en.wikipedia.org/wiki/Taylor%27s_theorem#Estimates_for_the_remainder

@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.