Inequality (1 Viewer)

[Vampire]

I'm having trouble with this question....


Prove:
a^2 + b^2 + c^2 => 3(abc)^2/3


Any help is appreciated...

 

[withoutaface]

Cubing both sides would probably work.
EDIT:a^6+b^6+c^6+2a^4b^2+2a^2b^4+6(abc)^2+2b^4c^2+2b^2c^4+2a^4c^2+2a^2c^4>=9(abc)^2
a^6+b^6+c^6+2a^4b^2+2a^2b^4+2b^4c^2+2b^2c^4+2a^4c^2+2a^2c^4>=3(abc)^2
On second thoughts it looks a bit messy:/

 

[turtle_2468]

Apply AM-GM to the three terms on the left.

 

[Trev]

Apply AM-GM to the three terms on the left.

What is AM-GM?

 

[shafqat]

arithmetic mean >= geometric mean
but can we assume the result in 4u david? i was under the impression it had to be proved, as most of the inequalities are actually just variations on am-gm

 

[gordo]

u assume the result
then u work backwards to prove it...

 

[turtle_2468]

hey shafqat,
I think you're probably right... usually you can't assume it.
But I guess in the particular case of n=3 it's quite hard (I think most ppl, myself included, wouldn't work it out), so it's assumable... esp as you need to see that a^2 etc are the terms you sub in...

Here's a more formal proof.
1) Prove AM-GM for n=2 (ie x^2+y^2>=2xy)
2) Prove it for n=4 (ie x^4+y^4+w^4+z^4>=4xywz; you use step 1 a few times..)
3) the proper form of step 2 is actually x+y+w+z>=4(xywz)^(1/4). Hence using this form substitute z=(x+y+w)/3, cancel out a little and you get the desired result

In general, to prove AM-GM (you DO NOT NEED THIS PROOF for 4U HSC):
1) Prove for n=2
2) Prove for induction that n=2^k where k is an integer works
3) substitute the last term = the average of the other terms, cancel out to get 2^k-1
4) repeat step 3 to get desired number.

 

[shafqat]

great post
thats pretty much all you'll need to know for 4u inequalities guys