If non-zero real numbers b and c are such that min f(x) > max g(x), where f(x) = x² + 2bx + 2c² and g(x) = -x² - cx + b² (x ∈ R), then |c/b| lies in the interval: (0, 1/2), [1√2, √2], [1/2, 1√2], (√2, ∞)

f(x) = x² + 2bx + 2c² = x² + 2 × b × x + b² - b² + 2c² (Add and subtract b²) = (x + b)² - b² + 2c²
From the above expression, the minimum value of f(x) is -b² + 2c² (1) (minimum value of the square term is zero)
g(x) = -x² - cx + b² = -(x² + 2cx - b²) = -(x² + 2cx + c² - c² - b²) = -((x + c)² - c² - b²) = -(x + c)² + b² + c²
From the above expression, the maximum value of g(x) is b² + c² (2)
It is given in the question that min f(x) > max g(x)
Using (1) and (2) we get, -b² + 2c² > b² + c² ⇒ c² > 2b² ⇒ c²/b² > 2 ⇒ |c/b| > √2