Since x>= 0, we can square this inequality on both sides.
1 + x >= 2sqrt(x)
x^2 + 2x + 1 >= 4x
x^2 - 2x + 1 >= 0
(x-1)^2 >= 0
A square number always has to be non-negative.
i think the problem should state that this holds for all real x
I'm a bit curious, does the x >= 0 imply that x is a real number?
Since something like 3i can't be greater than 0 right?
yeah, inequalities with complex numbers aren't well defined lol
i guess the x >= 0 would imply that