## anonymous one year ago prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

1. anonymous

2. anonymous

i think you made an error typing it up; that's a false statement

3. anonymous

4. anonymous

:o i hope not. Wolfram alpha seems to agree with me https://www.wolframalpha.com/input/?i=2%E2%88%9A%28x%29+%2B+1%2F%E2%88%9A%28x+%2B+1%29+%3C%3D+2%E2%88%9A%28x%2B1%29

5. anonymous

i wonder if you can use Taylor's theorem

6. anonymous

actually it's just the mean-value theorem, even easier

7. anonymous

Taylor's Theorem? O.O should it be that complicated?

8. anonymous

$$0\le f(x+1)-f(x)\le f'(\xi)$$ by the fact it's increasing and by the mean-value theorem for some $$\xi\in(x,x+1)$$. we know $$f'(\xi)=\frac1{2\sqrt{\xi}}$$ and since $$\xi<x+1\implies \sqrt{\xi}<\sqrt{x+1} \implies \frac1{\sqrt{\xi}}>\frac1{\sqrt{x+1} }$$ hmm so the MVT bound is actually not narrow enough

9. anonymous

what is the function f in this case?

10. anonymous

wait ... so are you saying the MVT is inconclusive?

11. anonymous

I was thinking maybe show that 2√(x) + 1/√(x + 1) > 2√(x+1) for some x in [0,inf) is false

12. ganeshie8

You may use AM-GM inequality : $\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}$

13. anonymous

ok, so that gives 2sqrt(x+1) <= (2x+1)/sqrt(x)

14. ganeshie8

rearrange in useful way

15. anonymous

ok, so that gives 2sqrt(x) <= 2sqrt(x) + 1/sqrt(x) we're close :O

16. anonymous

2sqrt(x+1)*

17. ganeshie8

$\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}$ $2\sqrt{x}\sqrt{x+1}+1\le 2(x+1)$ divide $$\sqrt{x+1}$$ both sides and you're done!

18. anonymous

how did you get 2(x+1) on the right side though?

19. anonymous

is it the fact that (2x+1) <= 2x+2= 2(x+1)

20. ganeshie8

$\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}$ $2\sqrt{x}\sqrt{x+1}\le 2x+1$ add 1 to both sides $2\sqrt{x}\sqrt{x+1}+1\le 2(x+1)$ divide $$\sqrt{x+1}$$ both sides and you're done!

21. anonymous

:OOOOO GENIUS!!!

22. ganeshie8

if you don't happen to remember the AM-GM inequality, then here is an alternative let $$a=\sqrt{x}$$ and $$b=\sqrt{x+1}$$ then $$0\le (a-b)^2 \implies 2ab\le a^2+b^2$$ (this is actually a proof of AM-GM inequality)

23. anonymous

yeah, i knew the proof of AM-GM inequality but didn't think that it would be used for this problem

24. anonymous

thanks a bunch @ganeshie8 :')

25. ganeshie8

np:) trust that almost all the homework proofs/problems are simple and are based on the already proven results