Let x 1 ≠ x2, f(x 1) = f(x2), then x 1e x 1 = x2e x2.

Logarithm of both sides: x1+e x1= x2+e x2+e x2.

Because e x > x (x >: 0), so x1+e x1>; x 1 + x 1 = 2x 1，x2+e^x2 & gt； x2 + x2 = 2x2 .

To sum up: x1+e x1> 2x1,x2+e x2 > 2x2, and f(x 1) = f(x2), so 2x 1 = 2x2, so x 1 = x2, which contradicts x 1 ≠ x2.

Therefore, if x 1 ≠ x2 is known and f(x 1) = f(x2), then x1+x2 >; 2.