IMC 2009 · Day 2 · P7

We have $f''(x) - 2 f'(x) - 3 (f'(x) - 2 f(x)) \ge 0$, $x \in [0, \infty)$.

Let $g(x) = f'(x) - 2 f(x)$, $x \in [0, \infty)$. It follows that $$g'(x) - 3 g(x) \ge 0, \quad x \in [0, \infty),$$ hence $$(g(x) e^{-3x})' \ge 0, \quad x \in [0, \infty),$$ therefore $$g(x) e^{-3x} \ge g(0) = -2, \quad x \in [0, \infty) \quad \text{or equivalently}$$ $$f'(x) - 2 f(x) \ge -2 e^{3x}, \quad x \in [0, \infty).$$ Analogously we get $$(f(x) e^{-2x})' \ge -2 e^{x}, \quad x \in [0, \infty) \quad \text{or equivalently}$$ $$(f(x) e^{-2x} + 2 e^{x})' \ge 0, \quad x \in [0, \infty).$$ It follows that $$f(x) e^{-2x} + 2 e^{x} \ge f(0) + 2 = 3, \quad x \in [0, \infty) \quad \text{or equivalently}$$ $$f(x) \ge 3 e^{2x} - 2 e^{3x}, \quad x \in [0, \infty).$$