Since (a > 0) and (b > 0), we know that (b > 0), which implies that (-3b^2 < 0). Therefore, (\Delta < 0), which means that the quadratic expression (a^2 - ab + b^2 - 1) will always be greater than or equal to 0 when (a > 0) and (b > 0).
Therefore, we can conclude that (a^2 - ab + b^2 \geq 1) when (a > 0) and (b > 0).
To prove that (a^2 - ab + b^2 \geq 1) when (a > 0) and (b > 0), we can rewrite the inequality as:
(a^2 - ab + b^2 - 1 \geq 0)
Now, let's consider the expression (a^2 - ab + b^2 - 1) as a quadratic in terms of (a). The discriminant of this quadratic expression is given by:
(\Delta = (-b)^2 - 4(1)(b^2 - 1) = b^2 - 4b^2 + 4 = -3b^2 + 4)
Since (a > 0) and (b > 0), we know that (b > 0), which implies that (-3b^2 < 0). Therefore, (\Delta < 0), which means that the quadratic expression (a^2 - ab + b^2 - 1) will always be greater than or equal to 0 when (a > 0) and (b > 0).
Therefore, we can conclude that (a^2 - ab + b^2 \geq 1) when (a > 0) and (b > 0).