how do you do this inequality? (1 Viewer)

[derek_]

show that (a+b+c)/3 is greater or equal too the cube root of (abc)

it's in the patel book but pity they don't have worked solutions ... i've tried assuming it's true but i didn't go any where as far as :

(a+b+c)^3/27 > or equal too (abc)

lol ._.

 

[Slidey]

arithmetic-geometric

 

[Premus]

so can we just state...using arithmetic geometric inequalities , i got this and this ?

Thanks

 

[derek_]

after looking at page 306 in the patel book, i think you've actually got too go through the entire gruesome process of it all ._.

 

[Estel]

saying "AM-GM" does not constitute a proof

 

[Idyll]

To do it, first you need to establish that a^2 + b^2 + c^2 - ab-bc-ac>=0

Then multiply both sides by a+b+c

Then make an appropriate substitution.

 

[Jase]

start of with the basic basic thing:

a^2 + b^2 >= 2ab
similarly: b^2 + c^2 >= 2bc and a^2 + c^2 >= 2ac

add them all: 2(a^2 + b^2 + c^2) >= 2(ab + bc+ ac)
now multiply both sides by ( a + b + c)

a^3 + ab^2 + bc^2 + ............. + c^3 >= a^2 b + a^2 c ............ + abc + abc + abc
hence a^3 + b^3 + c^3 >= 3abc

now sub a-> a^1/3 and b ...e.t.c

 

[Veck]

someone explain this AM-GM thing to me quickly, I've never heard of it..... and it seems to be your bible..... hmm.... or maybe it's some joke I never got

 

[Slidey]

so can we just state...using arithmetic geometric inequalities , i got this and this ?

Thanks

Err. If you're asked to prove an inequality, OBVIOUSLY, just saying a few words won't suffice. However, in the case of using the arithmetic-geometric mean inquality to prove something, I was under the impression that was allowed.

What Jase has done is simply used the AM-GM on a^2 and b^2:

(a^2+b^2)/2 >= sqrt(a^2.b^2), i.e.:
a^2+b^2 >= 2ab

This is what I meant when I said "arithmetic-geometric". While what you are trying to prove IS simply a version arithmetic-geometric mean inequality, I am pretty sure you can use the basic version to prove it.

Veck: what's the average of a and b? (a+b)/2. That's arithmetic mean. It's got to do with progression and such. Likewise with geometric mean - sqrt(ab) is its counterpart for geometric progression. Let's not forget harmonic progression either.

A pretty well-known inequality is that the arithmetic mean is greater than or equal to the geometric mean. i.e.:
(a+b)/2 >= sqrt(ab)