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inequalities questions.. (1 Viewer)

who_loves_maths

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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.
 

Slidey

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What, no Ph.B? I'm dissapointed.
 

KFunk

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haboozin said:
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)

&ge; (Rxyz)<sup>1/4</sup> ... from (1)

Hence R<sup>3/4</sup> &ge; (xyz)<sup>1/4</sup> so (x+y+z)/3 &ge; (xyz)<sup>1/3</sup>

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

who_loves_maths

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^ 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.
 

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