Am>gm (1 Viewer)

untouchablecuz · Oct 18, 2009

jetblack2007 said:
Yeah, I agree. It's a well known mathematical fact. Just say 'Using the AM-GM inequality...'

even then, if you lose a mark for it, it's 1 mark out of 120. I accidentally skipped a page, lost 7 marks or something, and still got 95.

you must of killed it

kwabon · Oct 18, 2009

untouchablecuz said:
you must of killed it

but last year had a really really low cut off mark for a band 6. but dont get me wrong 95 is amazing.

kwabon · Oct 18, 2009

untouchablecuz said:
i would expect a "hence or otherwise"

if it does say that, shotty your method.

GUSSSSSSSSSSSSS · Oct 18, 2009

untouchablecuz said:
$Show\ x\leq e^{x-1}\ (calculus \ or \ graphs)\\A=\frac{a_{1}+a_{2}+...+a_{n}}{n}\\\frac{a_{k}}{A}\leq e^{\frac{a_{k}}{A}-1}\Rightarrow \frac{a_{1}}{A}\frac{a_{2}}{A}...\frac{a_{n}}{A}\leq e^{\frac{a_{1}}{A}-1+\frac{a_{2}}{A}-1+...+\frac{a_{n}}{A}-1}\\But \ \frac{a_{1}}{A}-1+\frac{a_{2}}{A}-1+...+\frac{a_{n}}{A}-1=(a_{1}+a_{2}+...+a_{n})\frac{1}{A}-n=\frac{nA}{A}-n=0\\\therefore \frac{a_{1}a_{2}...a_{n}}{A^n}\leq 1\Rightarrow a_{1}a_{2}...a_{n}\leq A^n\Rightarrow \frac{a_{1}+a_{2}+...+a_{n}}{n}\geq (a_{1}a_{2}...a_{n})^{\frac{1}{n}}$

i suggest you commit this to memory

WOW man i wud never have come up with something like that,,,,,hence proving that it IS HARD for me to prove, maybe not for you xD ty for doin it so easy =]]]

jet · Oct 18, 2009

GUSSSSSSSSSSSSS said:
WOW man i wud never have come up with something like that,,,,,hence proving that it IS HARD for me to prove, maybe not for you xD ty for doin it so easy =]]]

There's a few different proofs on Wikipedia; pick your favourite.

kwabon · Oct 18, 2009

jetblack2007 said:
There's a few different proofs on Wikipedia; pick your favourite.

LOLOLOL
must spread rep before giving to jetblack.

GUSSSSSSSSSSSSS · Oct 18, 2009

jetblack2007 said:
There's a few different proofs on Wikipedia; pick your favourite.

hahaha if only i cud open wikipedia during the exam =P

untouchablecuz · Oct 18, 2009

just to clarify guys; thats cambridge's proof

i dont wanna be sued for intellectual property theft or something

addikaye03 · Oct 18, 2009

http://www.artofproblemsolving.com/Wiki/index.php/Proofs_of_AM-GM

eldore44 · Oct 18, 2009

untouchablecuz said:
$Show\ x\leq e^{x-1}\ (calculus \ or \ graphs)\\A=\frac{a_{1}+a_{2}+...+a_{n}}{n}\\\frac{a_{k}}{A}\leq e^{\frac{a_{k}}{A}-1}\Rightarrow \frac{a_{1}}{A}\frac{a_{2}}{A}...\frac{a_{n}}{A}\leq e^{\frac{a_{1}}{A}-1+\frac{a_{2}}{A}-1+...+\frac{a_{n}}{A}-1}\\But \ \frac{a_{1}}{A}-1+\frac{a_{2}}{A}-1+...+\frac{a_{n}}{A}-1=(a_{1}+a_{2}+...+a_{n})\frac{1}{A}-n=\frac{nA}{A}-n=0\\\therefore \frac{a_{1}a_{2}...a_{n}}{A^n}\leq 1\Rightarrow a_{1}a_{2}...a_{n}\leq A^n\Rightarrow \frac{a_{1}+a_{2}+...+a_{n}}{n}\geq (a_{1}a_{2}...a_{n})^{\frac{1}{n}}$

i suggest you commit this to memory

at the risk of sounding like an idiot: is this proving x <= e^(x-1) assuming that (a1+a2...+an)/n <= nth root of (aXbX..Xn)

jet · Oct 18, 2009

eldore44 said:
at the risk of sounding like an idiot: is this proving x <= e^(x-1) assuming that (a1+a2...+an)/n <= nth root of (aXbX..Xn)

He's just using it to help prove the AM-GM inequality; it's a means to an end.

untouchablecuz · Oct 18, 2009

just a Q

is proving x>=e^x-1 using graphs legit?

jet · Oct 18, 2009

Just do it this way:
f(x) = e^(x - 1) - x
f'(x) has one stationary point - it's a minimum.
At that minimum, f(x) = 0
Therefore f(x) ≥ 0

buchanan · Oct 18, 2009

untouchablecuz said:
just to clarify guys; thats cambridge's proof

i dont wanna be sued for intellectual property theft or something

To clarify even further, it should be said that it's Pólya's proof, not Cambridge's.

And it came to him in a dream!

untouchablecuz · Oct 18, 2009

buchanan said:
To clarify even further, it should be said that it's Pólya's proof, not Cambridge's.

And it came to him in a dream!

oo, indeed it is

on behalf of Mr. George Pólya estate, i am suing Cambridge for intellectual property theft

untouchablecuz · Oct 18, 2009

jetblack2007 said:
Just do it this way:
f(x) = e^(x - 1) - x
f'(x) has one stationary point - it's a minimum.
At that minimum, f(x) = 0
Therefore f(x) ≥ 0

yer but im asking that if i drew the graphs and stated that "from the graph of each function we can see that x<=e^{x - 1} for all real x", would it be accepted?

shaon0 · Oct 18, 2009

untouchablecuz said:
yer but im asking that if i drew the graphs and stated that "from the graph of each function we can see that x<=e^{x - 1} for all real x", would it be accepted?

I don't think so because your graph could be inaccurate. Best way to prove it is Jetblack2007's way.

duckcowhybrid · Oct 18, 2009

How do you people set out these fucked up inequalities. I have no clue how to set them out, I normally just work backwards till I get a perfect square > 0 or someshit then rewrite the whole thing forwards by saying consider (instert last line of working backwards).

jet · Oct 18, 2009

duckcowhybrid said:
How do you people set out these fucked up inequalities. I have no clue how to set them out, I normally just work backwards till I get a perfect square > 0 or someshit then rewrite the whole thing forwards by saying consider (instert last line of working backwards).

I can imagine where it will go in my head.
Also, practice. Generally you will have seen the proof/question before. If BOS think nobody will have or have approached the question from a different perspective, it will be a multi-part question leading you into the result.

untouchablecuz · Oct 18, 2009

duckcowhybrid said:
How do you people set out these fucked up inequalities. I have no clue how to set them out, I normally just work backwards till I get a perfect square > 0 or someshit then rewrite the whole thing forwards by saying consider (instert last line of working backwards).

do alot of them and it becomes natural

initially i was very much like this

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