Solve for Reals

Hmmm , 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 .

Similarly 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 ;