Reciprocal of a convex decreasing function:

Question

I have the following math problem that I haven't been able to solve. I would be very grateful of any idea or solution you can give me. Thanks!

Let $f(x): [0,\infty)\rightarrow (0,\infty)$ such that $f'(x)<0$ and $f''(x)>0$. What are the necessary and sufficient conditions for $1/f(x)$ to be concave.

I approach to this problem by using derivatives and got to the equation: $$2(f'(x))^2 -f''(x)f(x)\leq0$$ Yet I feel something smarter can be done that gives a more precise characterization.

Thanks!

Answer 1

$f'(x)<0,\ f''(x)>0$. Then $1/f$ is concave

Proof : $ 0<t<1$. Then $$ f(ta+(1-t)b)\leq tf(a) +(1-t)f(b) $$ so that $$\frac{1}{f(ta+(1-t)b) }\geq \frac{1}{tf(a) }+\frac{1}{(1-t)f(b) }\geq t\frac{1}{f(a)} + (1-t) \frac{1}{f(b)} $$

Answer 2

I think the accepted answer in this post is incorrect. The statement has a counterexample.

Let $f(x) = \frac1{(x+1)^2}$. Then $f'(x)=-2(x+1)^{-3} <0$ and $f''(x)=6(x+1)^{-4} >0$.

However, $\frac1{f(x)}= (x+1)^2$ is not concave.

Since the original question was asking about the necessary and sufficient condition, it should be the inequality in the OP: $$ 2(f'(x))^2 -f''(x)f(x)\leq0. $$