x = k + 1, k ∈ ℕ et y = g + 1 ∈ ℕ
xy = (k + 1) × (g + 1) = kg + k + g + 1
y = g + 1
xy - y = (kg + k + g + 1) - (g + 1) = kg + k
donc xy > y
x = k + 1
xy - x = (kg + k + g + 1) - (k + 1) = kg + g
donc xy > x
2) 2xy = 2kg + 2k + 2g + 2
x + y = k + g + 2
2xy - (x + 1) = 2kg + 2k + 2g + 1 - ( k + g + 2) = 2kg + k + g
si 2xy - (x + 1) ∈ ℝ*+ alors 2xy > x + y, or (2kg + k + g) ∈ ℝ*+ car k et g ∈ ℕ
