Proving d, e and f as Linear Combinations of a, b and c

[andreass]

Homework Statement

Given:
a>=b>=c>=0,
d>=e>=f>=0,
a>=d
a+b>=d+e
a+b+c=d+e+f
a,b,c,d,e,f belong to Real numbers

Prove that d, e, f can be expressed as linear combinations of a, b and c in such way:
d=(c1+c2)*a+(c3+c4)*b+(c5+c6)*c
e = (c1+c6)*a+(c2+c4)*b+(c3+c5)*c
f=(c1+c3)*a+(c2+c5)*b+(c4+c6)*c

c1, c2, c3, c4, c5, c5 >=0

The Attempt at a Solution

Only thing I can prove is that c1+c2+c3+c4+c5+c6 = 1 (using a+b+c=d+e+f).
I think I need to find some expression for b, to be able do something, but I'm not sure.

Any suggestions?

 

[Quinzio]

a+b+c=d+e+f

Is it correct or should it be:
a+b+c>=d+e+f ?

 

[andreass]

a+b+c=d+e+f
Is it correct ?

there's no mistake.
a+b+c=d+e+f is correct

 

[Quinzio]

I've tried to play with the expressions and something else is coming out, but I don't get nowhere.
Example:
may I let
a=d
b=c=e=f=0 ?
I don't see why I couldn't, every equation is respected.
This leads to
c1+c2=1
c1+c6=0
c1+c3=0

Since:
c1+c2=1 and c1+c2+c3+c4+c5+c6=1
c3+c4+c5+c6=0

So far so good.
Then I assume
d=e=f=1
a=(1+k)
b=1
c=(1-k)
again, everything is respected.
Using the last assumption into d=(c1+c2)*a+(c3+c4)*b+(c5+c6)*c
I get c1+c2=c5+c6
Since c1+c2=1, c5+c6=1

Then recalling c3+c4+c5+c6=0
I conclude c3+c4=-1

But this is against the problem constraint:
c1, c2, c3, c4, c5, c5 >=0

So what do we do here ?
Is there any flaw in my logic chain ? Any mistake ?
I don't get the final goal of the problem...
c1...c6 should be constant real number, or should they be functions of a,b,c,d,e,f ?

 

[andreass]

c1..c6 are coefficients, but I think they will probably change when a,b,c,d,e,f change.

 

[Quinzio]

c1..c6 are coefficients, but I think they will probably change when a,b,c,d,e,f change.

OK, so they are functions of a,b,c,d,e,f.

But what about my conclusion
c3+c4=-1
But this is against the problem constraint:
c1, c2, c3, c4, c5, c5 >=0

Do you have a solution, if not bring my conclusion to your teacher, whoever.