# Inequalities on convexity

Let $$f$$ be second-order differentiable on $$(-\infty,+\infty)$$, and $$|f(x)|^3$$ be convex on $$(-\infty,+\infty)$$. Proved that for any $$x$$, $$f(x)f''(x)+2[f'(x)]^2\geqslant0.$$ I try to deform the inequality,and got it $$f(x)f''(x)+(f'(x))^2\geqslant-(f'(x))^2\implies(f(x)f'(x))'\geqslant-(f'(x))^2$$ But then I don't know how to deal with it, and I don't know what "$$|f(x)|^3$$ be convex" can do...

• $(|f|^3)'' = (3f^2f')' = 6f(f')^2 + 3f^2f'' = 3f(2(f')^2 + ff'')$ on $\{f >0 \}$ may help ... – martini Jan 20 at 13:01
• @martini thank you ,Do we just need to discuss the positive and negative of $f$? – Hilbert1994 Jan 20 at 13:06

The idea is that either $$f(x) = 0$$ (in which case the inequality holds trivially) or $$|f(x)| = f(x)$$ or $$|f(x)| = -f(x)$$ in a neighbourhood of $$x$$ (in which case we can compute the second derivative).
Consider first the case that $$f(x_0) > 0$$. Then $$|f(x)|^3 = f(x)^3 > 0$$ for all $$x$$ in an open interval $$I$$ containing $$x_0$$. It follows that $$0 \le \frac{d^2}{dx^2} (f(x))^3 = 3f(x) \bigl(2f'(x)^2 + f(x) f''(x) \bigr)$$ for all $$x \in I$$, and in particular $$2f'(x_0)^2 + f(x_0) f''(x_0) \ge 0$$.
The case $$f(x_0) < 0$$ works similarly.
Finally, if $$f(x_0) = 0$$ then $$2f'(x_0)^2 + f(x_0) f''(x_0) = 2f'(x_0)^2 \ge 0$$.