Question

prove 2√(x) + 1/√(x + 1) <= 2√(x+1) for all x in [0,inf)

Answer 1

maybe by contradiction?

Answer 2

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

Answer 3

oh oops, i misread

Answer 4

: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

Answer 5

i wonder if you can use Taylor's theorem

Answer 6

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

Answer 7

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

Answer 8

$$0\le f(x+1)-f(x)\le f'(\xi)$$ 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

Answer 9

what is the function f in this case?

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

rearrange in useful way

Answer 15

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

we're close :O

Answer 16

2sqrt(x+1)*

Answer 17

\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\]
\[2\sqrt{x}\sqrt{x+1}+1\le 2(x+1)\]

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

Answer 18

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

Answer 19

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

Answer 20

\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\]

\[2\sqrt{x}\sqrt{x+1}\le 2x+1\]
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!

Answer 21

:OOOOO GENIUS!!!

Answer 22

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)

Answer 23

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

Answer 24

thanks a bunch @ganeshie8 :')

Answer 25

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