Le corrigé

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xyz = 4 (x+y+z) 0 < x 6y6z
x y z
(x, y, z) 0 < x 6y6z
x2z6xyz = 4 (x+y+z)64 (z+z+z) = 12z
x2z612z z z > 0x2612 x
x63x1 2 3
•x= 1 yz = 4 (1 + y+z) 1 6y6z
y
yz = 4 (1 + y+z)64 (1 + 2z)612z
16y612
z
yz = 4 + 4y+ 4z⇐⇒ (y−4) z= 4 + 4y
y−4 4 + 4y y −4 (4 + 4y)−
4 (y−4) = 20 y−4 20
y∈ {5; 6; 8; 9; 14; 20} ∩ [[1; 12]] = {5; 6; 8; 9}
y= 5 5z= 4 (6 + z)z= 24
y= 6 6z= 4 (7 + z)z= 14
y= 8 8z= 4 (9 + z)z= 9
y= 9 9z= 4 (10 + z)z= 8
z>y= 9
(1; 5; 24) (1; 6; 14) (1; 8; 9)
y(z−4) = 4+4z z
{5; 6; 8; 9}(1; 5; 5) ,(1; 5; 6) ,(1; 5; 8) ,(1; 5; 9) ,(1; 6; 6) , . . .

•x= 2 2yz = 4 (2 + y+z) 2 6y6z
2yz 612z y 66 (y−2) z= 4 + 2y
y−2|y−2
y−2|4+2yy−2|(4 + 2y−2 (y−2)) = 8 y∈ {3; 4; 6}
y= 3 3z= 2 (5 + z)z= 10
y= 4 4z= 2 (6 + z)z= 6
y= 6 6z= 2 (8 + z)z= 4
(2; 3; 10) (2; 4; 6)
•x= 3 3yz = 4 (3 + y+z) 3 6y6z
3yz 612z y 64 (3y−4) z= 12 + 4y
3y−4|12 + 4y
3y−4|3y−43y−4|(3 (12 + 4y)−4 (3y−4)) = 20
3y−4∈ {1; 2; 5; 10; 20}y∈ {2; 3; 8}y= 3
y= 3 9z= 4 (6 + z)
S={(1; 5; 24) ,(1; 6; 14) ,(1; 8; 9) ,(2; 3; 10) ,(2; 4; 6)}
3
3
n−1n n + 1
(n−1)3+n3+ (n+ 1)3=n3−3n2+ 3n−1 + n3+n3+ 3n2+ 3n+ 1 = 3n3+ 6n= 3nn2+ 2
3
n n+1 n+2
(a+b)3=a3+ 3a2b+ 3ab2+b3