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

  • Thread starter StonedPanda
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In summary, The conversation is about proving an inequality using the equation a^4 + b^4 + c^4 + d^4 = 4abcd and using w=(abc)^1/3/. The conversation also includes a suggestion to consider different situations and to rewrite the equation. Finally, a proof is provided for the inequality 3(jkl)^(1/3) <= j^3 + k^3 + l^3.
  • #1
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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  • #2
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.
 
  • #3
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
 
  • #4
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 dosen'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.
 
  • #5
I meant prove a^3+b^3+c^3>=3abc using w=(xyz)^(1/3)

and yes i did mean earlier >=
 
  • #6
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:
  • #7
i got it, will post proof tomorrow.
 
  • #8
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
 

1. What is the significance of the inequality a^3+b^3+c^3>=3abc?

The inequality a^3+b^3+c^3>=3abc is known as the AM-GM inequality, which stands for the Arithmetic Mean-Geometric Mean inequality. This inequality is an important concept in mathematics and is frequently used in various fields, including physics, economics, and engineering.

2. How can I prove the inequality a^3+b^3+c^3>=3abc?

The most common way to prove the AM-GM inequality is by using mathematical induction. This involves proving the inequality for a specific case, usually when a, b, and c are equal, and then showing that it holds true for all other cases using the principle of mathematical induction.

3. What are the applications of the inequality a^3+b^3+c^3>=3abc?

The AM-GM inequality has many real-world applications. In economics, it is used to find the most efficient distribution of resources. In physics, it is used to find the minimum energy required to perform a certain task. In geometry, it is used to prove the existence of certain geometric shapes.

4. Can the inequality a^3+b^3+c^3>=3abc be extended to more than three variables?

Yes, the AM-GM inequality can be extended to any number of variables. The general form of the inequality is a1^n+a2^n+...+an^n>=n√(a1*a2*...*an), where n is the number of variables. This is known as the n-th power mean inequality.

5. Are there any alternative ways to prove the inequality a^3+b^3+c^3>=3abc?

Yes, there are various other methods that can be used to prove the AM-GM inequality, such as the Cauchy-Schwarz inequality, the Jensen's inequality, and the rearrangement inequality. These methods may require a deeper understanding of advanced mathematical concepts, but they can provide alternative and sometimes more elegant proofs of the AM-GM inequality.

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