(1)   a + bcd = 2
(2)   b + acd = 2
(3)   c + abd = 2
(4)   d + abc = 2

Subtract eq (2) from eq (1)

a - b + bcd - acd = 0
a - b - acd + bcd = 0
  (a-b) - cd(a-b) = 0
      (a-b)(1-cd) = 0

Since a>b, a-b>0
so         1-cd = 0         
            -cd = -1
             cd = 1

Subtract eq (3) from eq (1)

a - c + bcd - abd = 0
a - c - abd + bcd = 0
  (a-c) - bd(a-c) = 0
      (a-c)(1-bd) = 0
Since a>c, a-c>0
so         1-bd = 0         
            -bd = -1
             bd = 1

Subtract eq (4) from eq (1)

a - d + bcd - abc = 0
a - d - abc + bcd = 0
  (a-d) - bc(a-d) = 0
      (a-d)(1-bc) = 0

Since a>d, a-d>0
so         1-bc = 0         
            -bc = -1
             bc = 1

So we have 

cd = 1
bd = 1
bc = 1

cd = bd = 1, so c = b
cd = bc = 1, so d = b

So b = c = d

a + bcd = 2
a + bbb = 2
 a + b³ = 2

b + acd = 2
b + abb = 2
b + ab² = 2

So we have the system
 a + b³ = 2
b + ab² = 2

Solve the 1st for a

a = 2 - b³

Substitute in the 2nd eq

     b + (2-b³)b² = 2
     b + 2b² - b5 = 2
-b5 + 2b² + b - 2 = 0
 b5 - 2b² - b + 2 = 0

Possible rational roots are ±1,±2

1|1 0 0 -2 -1  2
 |  1 1  1 -1 -2 
  1 1 1 -1 -2  0

(b-1)(b4+b³+b²-b-2)=0

1|1 1 1 -1 -2
 |  1 2  3  2
  1 2 3  2  0

(b-1)(b-1)(b³+2b²+3b+2)=0

-1|1  2  3  2
  |  -1 -1 -2 
   1  1  2  0

(b-1)(b-1)(b+1)(b²+b+2)=0

The last factor does not yield real
roots.

So the real roots are 1 and -1

b can't be 1 for

 a + b³ = 2
 a + 1³ = 2
  a + 1 = 2
      a = 1
which contradicts a>b

Therefore b=c=d=-1

    a + b³ = 2
 a + (-1)³ = 2
     a - 1 = 2
         a = 3

Edwin