inequalities questions.. (2 Viewers)

[Vampire]

hahaha what's that like - usyd law?

 

[who_loves_maths]

Originally Posted by 香港!
must've been some intense studying @_@
its been nearly 2 months lol
how do u do that? without coming online at all??
i wish i could be as focused as u

well, i did say "sort of"... certainly didn't study for two months straight... and what's the point anyway, i just want to get into the course i'm gunning for.

 

[香港!]

hahaha what's that like - usyd law?

i think its UNSW Med

 

[sikeveo]

another one of those

 

[who_loves_maths]

Originally Posted by Vampire
hahaha what's that like - usyd law?

how'd you know i was interested in law? hehe...

 

[Slidey]

What, no Ph.B? I'm dissapointed.

 

[香港!]

hey captain how did u go today?

 

[KFunk]

if they give you:

prove:

x^3 + y^3 + z^3 >= 3xyz


without any lead in (i know that is pretty farfetched.

If you prove the n=3 case for AM-GM, as has been suggested, it fall into place pretty quick. This is the method I've learnt for proving it, if anyone is interested:

First do the simple part and prove (x + y)/2 ≥ √(xy) ... (1)

Let R = (x+y+z)/3

R = (R + x + y + z)/4

≥ [√(RX) + √(yz)]/2 .... from (1)

≥ (Rxyz)^1/4 ... from (1)

Hence R^3/4 ≥ (xyz)^1/4 so (x+y+z)/3 ≥ (xyz)^1/3

Does anyone know an easier method or is this just the stock standard way of going about it?

 

[who_loves_maths]

^ actually, the "standard" way would be to use mathematical (forward) induction. but that is much more prolific than the method you employed.

backward induction via specialisation (i.e. what you did) is a very simple method of proving the general Cauchy's Inequality. the idea is that 1) if the case for n = k is true, then the case for n = (k -1) must also be true.
and 2) the two-variable AM-GM inequality is easily extended to all cases of the form n = 2^m ; where 'm' is positive integral (eg. you did it for n=4 where m=2).

and then combining both parts, the general inequality for all positive integral 'n' follows immediately since as m -> infinity, 2^m -> infinity as well.