Can some one help me out here?

I´m doing som maths repetion and I cant get some things right here.
Prove:

If A=(x+y+z)/3
and G=(xyz)^(1/3)

Then prove that G=
I can´t find the ****ing proof!!!
Any one here that can kil this bitch???

BTW, x,y,z are all >0

yeah well, life\'s a b1tch!

Originally posted by Bossieman
I´m doing som maths repetion and I cant get some things right here.
Prove:

If A=(x+y+z)/3
and G=(xyz)^(1/3)

Then prove that G=
I can´t find the ****ing proof!!!
Any one here that can kil this bitch???

not sure what you need for proof, but if we give each letter a numeric value, it all works out.

example;
x=1
y=2
z=3

A=(1+2+3)/3
G=(1x2x3)^(1/3)

A=6/3
G=6/3

A=2
G=2

this probably doesn\'t help at all, because you probably need a theory of some sorts for proof.

well one way i can think of this is;
A=(x+y+z)/3
G=cube root of (x*y*z)
now let x=y=z=1;
we get;
A=3/3=1
G=cube root of (1)=1
hence A=G

probably wrong:p but its been a while since i have done anything other than trig., integration and differnciation.

Originally posted by theomen

A=(1+2+3)/3
G=(1x2x3)^(1/3)

A=6/3
G=6/3

A=2
G=2

theomen 1/3 is not 3. It\'s 0\'3333333.....

so:
A= 2
G= 1,817

edit ~

E=mc sqaured

....

....

All your base are belong to us

AHAHAAHHAH What a ripper. Looks like a bit of multi variable mathematical induction.. if thats the proper term. The inductoin problems I\'m used to only have a single variable; which makes it a bit simpler. I had a bit of trouble trying to get the two damn equations to look like each other, so I\'ll conveniently change the sign from <= to < Hope u don\'t mind.

prove,
(xyz)^1/3 < (x+y+z)/3

assume x = 1, y = 2, z = 3,
(1.2.3)^1/3 < (1+2+3)/3
1.8 (2dp) < 2
therefore x = 1, y = 2, z = 3 is true

assume x = k, y = l, y = m is true
prove true for x = k+1, y = l+1, y = m+1,
((k+1)(l+1)(m+1))^1/3 < (k+1+l+1+m+1)/3
(klm+kl+km+k+lm+m+1)^1/3 < (k+l+m+3)/3
(k(lm+l+m+1)+l(m+1)+m+1)^1/3 < (k+l+m)/3 + 1  

disregard 1, since it becomes insignficant as k,l,m --> infinity

(k(lm+l+m+1)+l(m+1)+m+1)^1/3 < (k+l+m)/3
3.(k(lm+l+m+1)+l(m+1)+m+1)^1/3 < k+l+m

and from that you can see that for,
A = (x+y+z)/3
G = (xyz)^1/3
G < A
for x >= 1, y >=2, z >=3
or whatever......

Its currently 2:03am and I am very sleepy, so I might have boobed somewhere.

HAHAHAhA, I\'m such an idiot. It was just a matter of simple eyeballing; for A = G. Lewk!

(xyz)^1/3 = (x+y+z)/3
cube both sides
(xyz) = (x+y+z)^3/27
27(xyz) = (x+y+z)^3

so then u realise if each variable was 1
the left hand side would be 27 x 1
the right hand side would be 3^3 which is 27!!
SO IT WORKS! if x=y=z=1 !!!
Lewk,

(1.1.1)^1/3 = (1+1+1)/3

but if each variable has to unique.... the question is ****ed....

Well A is an average of x, y, and z, so you just have to prove the cube root of them is always equal to or less than their average.

Maybe that helps?

Obviously, if the variables are all the same, the average and cube root are equal so A = G.  So it\'s just a matter of proving that if they\'re different (2, 4, 6) the average (4) is always less than the cube root (3.6) Hmm... nevermind.

-Eik

Well, since it was proving that the average (4) is greater than the cube root (3.6), and not the other way around, that sounds good so far.

o shit, i really shoudn\'t be doing higher math nxt year

Originally posted by Adan


theomen 1/3 is not 3. It\'s 0\'3333333.....

so:
A= 2
G= 1,817

I know 1/3 isn\'t 3, but if you multiply by 1/3 is the same as dividing by 3.

6*1/3= 2
6/3 =2

instead of six times one third, think of it as six times one, over three.

6*(1/3)= (6*1)/3

Ah yes.. but 6*1/3 is not 6^1/3, the hat (^) denotes power.
So my name should really be Avatarr^, cuz I have POWER :evil:

I hate this problem, I dont know how to break it down