## ParthKohli one year ago Cool solution.

1. ParthKohli

Problem: Prove that$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e$for distinct reals $$a,b$$.

2. freckles

How many seconds can I have to think about it before you give the solution? :p

3. ParthKohli

4. freckles

$(\frac{1+a}{1+b})^{\frac{1}{a-b}} \\ =(\frac{1+b+a-b}{1+b})^\frac{1}{a-b} \\ =(1+\frac{a-b}{1+b})^{\frac{1}{a-b}}$ $=(1+\frac{1}{\frac{1+b}{a-b}})^\frac{1}{a-b} \\ =(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^\frac{1}{a-b}$ $\lim_{a-b \rightarrow \infty}(1+\frac{\frac{1}{1+b}}{\frac{1}{a-b}})^{\frac{1}{a-b}}=\lim_{a-b \rightarrow \infty} e^{\frac{1}{1+b}} \\ ...$ 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

5. ParthKohli

Ganeshie is here - he'll easily crack this one. :|

6. freckles

I did something wrong

7. freckles

I should have said 1/(a-b) goes to infinity

8. freckles

which would mean a-b goes to 0

9. ParthKohli

ganeshie8 is typing a reply... Well, here we go.

10. anonymous

$$\left(\frac{1 + a}{1+b}\right)^{\frac{1}{a-b}}< e\\\frac1{a-b}\log\left(\frac{1+a}{1+b}\right)<1\\\frac1{a-b}\left(\log(1+a)-\log(1+b)\right)<1\\\log(1+a)-\log(1+b)<(1+a)-(1+b)$$... is just a restatement of the fact that $$\log$$ is concave by Jensen's inequality

11. anonymous

all those steps are reversible, so the proof just follows from the fact that $$\log$$ is concave via Jensen's inequality

12. freckles

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 for any real b

13. ganeshie8

$e^{x-y} \gt 1+(x-y)+(x-y)^2/2 \gt 1+x-y = 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}$

14. anonymous

haha that one is great

15. anonymous

but hmm @ganeshie8 both you and i have only proven it for $$a-b>0$$ technically

16. ganeshie8

Ah right, I see..

17. anonymous

unless you can rigorously show that the even terms outweigh the odd ones in the expansion $$e^x=1+x+\frac12 x^2+\dots$$

18. anonymous
19. ganeshie8

looks the taylor series one also requires $$y\gt 0$$ $e^{x-y} \gt 1+(x-y)\gt 1+\dfrac{x-y}{1+y} = \frac{1+x}{1+y}$

20. anonymous

21. ParthKohli

@ganeshie8 That's exactly my answer. I think I got the incomplete question.

22. ParthKohli

Hey, your next project is to implement LaTeX in the drawing tool. That looks good.

23. ganeshie8

looks the inequality is true only when $$a,b$$ have same signs |dw:1437673377179:dw|

24. ParthKohli

So yes, I did get an incomplete question. You can't expect much from Facebook math groups.

25. ganeshie8

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