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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...

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  • $\begingroup$ $(|f|^3)'' = (3f^2f')' = 6f(f')^2 + 3f^2f'' = 3f(2(f')^2 + ff'')$ on $\{f >0 \}$ may help ... $\endgroup$ – martini Jan 20 at 13:01
  • $\begingroup$ @martini thank you ,Do we just need to discuss the positive and negative of $f$? $\endgroup$ – Hilbert1994 Jan 20 at 13:06
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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 $2f'(x_0)^2 + f(x_0) f''(x_0) = 2f'(x_0)^2 \ge 0$.

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