How can I prove a^3+b^3+c^3>=3abc using w=(abc)^1/3/?

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The discussion centers on proving the inequality a^3 + b^3 + c^3 ≥ 3abc using the formula w = (abc)^(1/3). Participants clarify that the inequality must be proven for positive numbers only, as it does not hold for negative values. One user suggests breaking down the proof into cases based on the values of a, b, and c, particularly when they are less than or greater than 1. The conversation also references the established equation a^4 + b^4 + c^4 + d^4 = 4abcd as a foundational concept for the proof.

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StonedPanda
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So I've been thinking about this for a while for an analysis class. I proved that a^4 + b^4 + c^4 + d^4 = 4abcd. Now I'm supposed to prove the inequality above using w=(abc)^1/3/. I'm not asking anyone to do my homework for me, but maybe someone could point me in the right direction?

Nice to be back here.
 
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i realize i might have posted this in the wrong place. I'm sorry. please move it, although i do think that it is a nice topic for this section.
 
Although this is slightly off-topic I don't how:

a^4 + b^4 + c^4 + d^4 === 4abcd

Let:
a = 1
b = 2
c = 4
d = 7

1^4 + 2^4 + 4^4 + 7^4 = 2674

However:
4*1*2*4*7 = 224

I'm confused :S
 
StonedPanda said:
So I've been thinking about this for a while for an analysis class. I proved that a^4 + b^4 + c^4 + d^4 = 4abcd. Now I'm supposed to prove the inequality above using w=(abc)^1/3/. I'm not asking anyone to do my homework for me, but maybe someone could point me in the right direction?

Nice to be back here.
This inequality doesn't hold for a=-1,b=-1,c=1 so I assume that it must be proven for positive numbers only? About the other response, I assume that you meant to type ">= 4abcd"; is this correct? My suggestion is to consider separately the situation where the numbers are less than 1 from what happens as the numbers a,b,c,d become greater than 1. Also what if you rewrote the equation so that one side would be a more definite quantity and still managed to use w somehow. Just some thoughts.
 
I meant prove a^3+b^3+c^3>=3abc using w=(xyz)^(1/3)

and yes i did mean earlier >=
 
argh i meant using a^4 + b^4 + c^4 + d^4 = 4abcd and d=(abc)^(1/3) prove a^3 + b^3 + c^3 + d^3 = 3abc for positive numbers a,b,c
 
Last edited:
i got it, will post proof tomorrow.
 
4xyzw<=x^4+y^4+z^4+w^4
(((( w=(xyz)^(1/3) ))))
4(xyz)^(4/3)=x^4+y^4+z^4+(xyz)^(4/3)
3(xyz)^(4/3)=x^4+y^4+z^4
((((a,b,c=x^4,y^4,z^4 respetively))))
3(abc)^(1/3)<= a + b + c
and now use a,b,c = j^3,k^3,l^3
 

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