Am>gm (1 Viewer)

Mileyjoe · Oct 17, 2009

AM i allowed to just state it?? or do i have to prove it or shizznit??

untouchablecuz · Oct 17, 2009

u gotta prove it

three line proof:

$(\sqrt{a}-\sqrt{b})^2\geq 0\\a-2\sqrt{ab}+b\geq 0\\\frac{a+b}{2}\geq \sqrt{ab}$

SparkyZ · Oct 17, 2009

untouchablecuz said:
u gotta prove it

three line proof:

$(\sqrt{a}-\sqrt{b})^2\geq 0\\a-2\sqrt{ab}+b\geq 0\\\frac{a+b}{2}\geq \sqrt{ab}$

I love you

GUSSSSSSSSSSSSS · Oct 17, 2009

untouchablecuz said:
u gotta prove it

three line proof:

$(\sqrt{a}-\sqrt{b})^2\geq 0\\a-2\sqrt{ab}+b\geq 0\\\frac{a+b}{2}\geq \sqrt{ab}$

but you only proved it for TWO objects =S

untouchablecuz · Oct 17, 2009

$Since, \frac{a+b}{2}\geq \sqrt{ab}\\\\\Rightarrow \frac{a+b}{4}\geq \frac{1}{2}\sqrt{ab}\\\Rightarrow \frac{c+d}{4}\geq\frac{1}{2}\sqrt{cd}\\\therefore \frac{a+b+c+d}{4}\geq \frac{1}{2}(\sqrt{ab}+\sqrt{cd})\\(\sqrt[4]{ab}-\sqrt[4]{cd})^2=\sqrt{ab}-2\sqrt[4]{abcd}+\sqrt{cd}\geq 0\Rightarrow \sqrt{ab}+\sqrt{cd}\geq2\sqrt[4]{abcd}\\\therefore \frac{a+b+c+d}{4}\geq \frac{1}{2}(\sqrt{ab}+\sqrt{cd})\geq \frac{1}{2}(2\sqrt[4]{abcd})=\sqrt[4]{abcd}$

heres for four variables

to prove for 3 variables, let d=(a+b+c)/3 and do some work

look in cambridge P 242 for a quick general proof

GUSSSSSSSSSSSSS · Oct 17, 2009

untouchablecuz said:
$Since, \frac{a+b}{2}\geq \sqrt{ab}\\\\\Rightarrow \frac{a+b}{4}\geq \frac{1}{2}\sqrt{ab}\\\Rightarrow \frac{c+d}{4}\geq\frac{1}{2}\sqrt{cd}\\\therefore \frac{a+b+c+d}{4}\geq \frac{1}{2}(\sqrt{ab}+\sqrt{cd})\\(\sqrt[4]{ab}-\sqrt[4]{cd})^2=\sqrt{ab}-2\sqrt[4]{abcd}+\sqrt{cd}\geq 0\Rightarrow \sqrt{ab}+\sqrt{cd}\geq2\sqrt[4]{abcd}\\\therefore \frac{a+b+c+d}{4}\geq \frac{1}{2}(\sqrt{ab}+\sqrt{cd})\geq \frac{1}{2}(2\sqrt[4]{abcd})=\sqrt[4]{abcd}$

heres for four variables

to prove for 3 variables, let d=(a+b+c)/3 and do some work

look in cambridge P 242 for a quick general proof

yeayea any EVEN number of objects is easy to prove, just using that first 3 line proof

however for ODD numbers it much harder, even proving it for 3 numbers takes quite a few lines

and then proving it for "n" numbers is FUCKEN hard (well at least I found it pretty damn hard)

so yea i think more clarification is needed on whether you can assume it's true or not =S

GUSSSSSSSSSSSSS · Oct 17, 2009

oh and i aint got cambridge so i cant check out that proof =[

Sarebs · Oct 17, 2009

Hey, sorry to hijack, but is GM included in 2U?

shaon0 · Oct 17, 2009

GUSSSSSSSSSSSSS said:
yeayea any EVEN number of objects is easy to prove, just using that first 3 line proof

however for ODD numbers it much harder, even proving it for 3 numbers takes quite a few lines

and then proving it for "n" numbers is FUCKEN hard (well at least I found it pretty damn hard)

so yea i think more clarification is needed on whether you can assume it's true or not =S

It's not that difficult. Conjecture what you think it might be and use induction.

GUSSSSSSSSSSSSS · Oct 17, 2009

shaon0 said:
It's not that difficult. Conjecture what you think it might be and use induction.

haha yea i did use induction, but it still took me quite a while =S well for ME it was hard lol maybe not for some of you xD

untouchablecuz · Oct 17, 2009

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

i suggest you commit this to memory

untouchablecuz · Oct 18, 2009

$\frac{a+b+c+d}{4}\geq \sqrt[4]{abcd}\\Let \ d=\frac{a+b+c}{3}\\\therefore \frac{a+b+c}{3}\geq \sqrt[4]{\frac{a^2bc+b^2ac+c^2ba}{4}}$

can someone help me out, im having a mindblock; i've forgetten how to deal with the a²bc+b²ac+c²ba term

shaon0 · 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

Nice, that's pretty good.

jet · Oct 18, 2009

untouchablecuz said:
$\frac{a+b+c+d}{4}\geq \sqrt[4]{abcd}\\Let \ d=\frac{a+b+c}{3}\\\therefore \frac{a+b+c}{3}\geq \sqrt[4]{\frac{a^2bc+b^2ac+c^2ba}{4}}$

can someone help me out, im having a mindblock; i've forgetten how to deal with the a²bc+b²ac+c²ba term

$\text{Let } a + b + c = 3d \text{ then } a + b + c + d = 4d \\ \text{From above, we have } a + b + c + d \geq 4\sqrt[4]{abcd} \\ 4d \geq 4\sqrt[4]{abcd} \\ d^4 \geq abcd \\ d^{3} \geq abc \\ d \geq \sqrt[3]{abc} \\ \text{Therefore } \frac{a + b + c}{3} \geq \sqrt[3]{abc}$

untouchablecuz · Oct 18, 2009

thanks mate

jet · Oct 18, 2009

There's another way which I was taught, but it's a bit longer

$\text{we know that } \left (x - y \right )^2 \geq 0 \\ \therefore x^2 + y^2 \geq xy \\ \text{Using this, it is easy to prove that } x^2 + y^2 + z^2 \geq xy + xz + yz \\ \text{Multiplying by } x + y + z, \left (x^2 + y^2 + z^2 \right )\left (x + y + z \right ) \geq \lef (xy + xz + yz)\left (x + y + z \right ) \\ x^3 + xy^2 + xz^2 + x^2y + y^3 + yz^2 + xz^2 + yz^2 + z^3 \geq x^2y + x^2z + xyz + xy^2 + xyz + y^2z + xyz + xz^2 + yz^2 \\ x^3 + y^3 + z^3 \geq 3xyz \\ \text{Letting } a = x^3, b = y^3, c = z^3 \text{ we have } \frac{a + b + c}{3} \geq \sqrt[3]{abc}$

untouchablecuz · Oct 18, 2009

you can also work backwards from this factorisation

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\a^2+b^2+c^2\geq ab+bc+ca\\\therefore a^3+b^3+c^3-3abc\geq ((ab+bc+ca)-(ab+bc+ca))=0\\a^3+b^3+c^3 \geq 3abc\\ Letting \ x=a^3, y=b^3 and \ z=c^3, \frac{x+y+z}{3}\geq \sqrt[3]{xyz}$

this is a good thread, everyone give your AM>GM tips

kwabon · Oct 18, 2009

um guys, i dont think you have to prove Am>Gm, well you dont have to prove it for n=2,
you can just state it. asked HSC marker and said it was alright to just state it.

it wouldnt hurt to prove it anywho, just three lines.

and btw untouchablecuz, pretty cool way to prove for n, but it is pointless putting that to memory since if inequalities do come up and ask you to prove for n number of terms, you will have to use previous parts of that particular question to prove it, and it always differs from year to year.

jet · Oct 18, 2009

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.

untouchablecuz · Oct 18, 2009

i would expect a "hence or otherwise"

Am>gm (1 Viewer)

New Member

Active Member

New Member

Active Member

Active Member

Active Member

Active Member

Member

...

Active Member

Active Member

Active Member

...

Banned

Active Member

Banned

Active Member

Banned

Banned

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)