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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe by contradiction?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think you made an error typing it up; that's a false statement

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0: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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i wonder if you can use Taylor's theorem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0actually it's just the meanvalue theorem, even easier

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Taylor's Theorem? O.O should it be that complicated?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$0\le f(x+1)f(x)\le f'(\xi)$$ by the fact it's increasing and by the meanvalue 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is the function f in this case?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wait ... so are you saying the MVT is inconclusive?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking maybe show that 2√(x) + 1/√(x + 1) > 2√(x+1) for some x in [0,inf) is false

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2You may use AMGM inequality : \[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so that gives 2sqrt(x+1) <= (2x+1)/sqrt(x)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2rearrange in useful way

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, so that gives 2sqrt(x) <= 2sqrt(x) + 1/sqrt(x) we're close :O

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\] \[2\sqrt{x}\sqrt{x+1}+1\le 2(x+1)\] divide \(\sqrt{x+1}\) both sides and you're done!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get 2(x+1) on the right side though?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0is it the fact that (2x+1) <= 2x+2= 2(x+1)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\] \[2\sqrt{x}\sqrt{x+1}\le 2x+1\] 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!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2if you don't happen to remember the AMGM inequality, then here is an alternative let \(a=\sqrt{x}\) and \(b=\sqrt{x+1}\) then \(0\le (ab)^2 \implies 2ab\le a^2+b^2\) (this is actually a proof of AMGM inequality)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, i knew the proof of AMGM inequality but didn't think that it would be used for this problem

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks a bunch @ganeshie8 :')

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2np:) trust that almost all the homework proofs/problems are simple and are based on the already proven results
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