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

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

Mathematics
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maybe by contradiction?
i think you made an error typing it up; that's a false statement
oh oops, i misread

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: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
i wonder if you can use Taylor's theorem
actually it's just the mean-value theorem, even easier
Taylor's Theorem? O.O should it be that complicated?
$$0\le f(x+1)-f(x)\le f'(\xi)$$ 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 \(\xi\frac1{\sqrt{x+1} }\) hmm so the MVT bound is actually not narrow enough
what is the function f in this case?
wait ... so are you saying the MVT is inconclusive?
I was thinking maybe show that 2√(x) + 1/√(x + 1) > 2√(x+1) for some x in [0,inf) is false
You may use AM-GM inequality : \[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\]
ok, so that gives 2sqrt(x+1) <= (2x+1)/sqrt(x)
rearrange in useful way
ok, so that gives 2sqrt(x) <= 2sqrt(x) + 1/sqrt(x) we're close :O
2sqrt(x+1)*
\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\] \[2\sqrt{x}\sqrt{x+1}+1\le 2(x+1)\] divide \(\sqrt{x+1}\) both sides and you're done!
how did you get 2(x+1) on the right side though?
is it the fact that (2x+1) <= 2x+2= 2(x+1)
\[\sqrt{x}\sqrt{x+1}\le \dfrac{2x+1}{2}\] \[2\sqrt{x}\sqrt{x+1}\le 2x+1\] 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!
:OOOOO GENIUS!!!
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)
yeah, i knew the proof of AM-GM inequality but didn't think that it would be used for this problem
thanks a bunch @ganeshie8 :')
np:) trust that almost all the homework proofs/problems are simple and are based on the already proven results

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