- #1

StonedPanda

- 60

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Nice to be back here.

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- Thread starter StonedPanda
- Start date

- #1

StonedPanda

- 60

- 0

Nice to be back here.

- #2

StonedPanda

- 60

- 0

- #3

tomfitzyuk

- 15

- 0

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

ramsey2879

- 841

- 0

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.StonedPanda said:

Nice to be back here.

- #5

StonedPanda

- 60

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I meant prove a^3+b^3+c^3>=3abc using w=(xyz)^(1/3)

and yes i did mean earlier >=

and yes i did mean earlier >=

- #6

StonedPanda

- 60

- 0

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

StonedPanda

- 60

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i got it, will post proof tomorrow.

- #8

StonedPanda

- 60

- 0

(((( 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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