3a=(b+c+d)^3
3b=(e+c+d)^3
3c=(e+a+d)^3
3d=(e+a+b)^3
3e=(c+a+b)^3.
find all reals (a,b,c,d,e).
Source - INMO.
1Hmmm , seems a little tricky -
Let us assume " a " is the greatest of them all .
So , a + e + d ≥ c + d + e ;
Consequently , c = ( a + e + d ) 33 ≥ ( c + d + e ) 33 = b
Also , since a ≥ d
Hence , e = ( a + b + c ) 33 ≥ ( b + c + d ) 33 = a
So we finally arrive at the inequality , e ≥ a , which clearly disproves our initial conjecture that " a " is the greatest one .
So the only possibility is that , e = a .
Again , since a ≥ b , so e ≥ b . b = ( c + d + e ) 33 ≥ ( c + d + b ) 33 = a
Finally , we get a = b also .
1Similarly proceeding , we may arrive at the following result -
a = b = c = d = e
for the possibility of such a system of equations to hold .
So , we need to solve - 3 a = ( 3 a ) 3
Giving rise to the solutions -
a = b = c = d = e = 0 ;
a = b = c = d = e = 13 ;
a = b = c = d = e = - 13 ;
