Relating the signs of $f$, $f'$ and $f''$ Let $f:\Bbb R \to \Bbb R$ be a function such that $f$ and $f'$ are differentiable for all $x$. If $f(x) \gt 0$ for all $x\ge 0$ and $f''(x)\lt 0$ for all $x\ge 0$, prove that $f'(x)\ge 0$ for all $x\ge 0$.
 A: If $f'(x_0)=a<0$, for some $a\ge 0$, the for all $x\ge x_0$:
$$
f'(x)-f'(x_0)=(x-x_0)f''(\eta), \,\,\text{for some $\eta\in(x_0,x)$},
$$
and thus
$$
f'(x)\lt f'(x_0)\,\,\,\text{for all $x\ge x_0$}.
$$
Thus
$$
f(x)=f(x_0)+\int_{x_0}^x f'(t)\,dt \le f(x_0)+a(x-x_0).
$$
But $f(x_0)+a(x-x_0)\to -\infty$, as $x\to\infty$, and hence $\lim_{x\to\infty}f(x)=-\infty$,
which contradicts the fact that $f(x)\ge 0$.
A: Suppose $f''(x)\le 0$ so that $f$ is concave.  If $f'(a)<0$ somewhere, then (since $f$ is concave) its graph is below the tangent line at $a$, and therefore $f$ is eventually negative.
A: Let's assume that $f'(a) < 0$ for some $a \geq 0$. Then since $f''(x) < 0$ for all $x \geq 0$, it follows that $f'(x) < f'(a) < 0$ for all $x > a$. It follows that $f(x)$ is strictly decreasing in $[a, \infty)$ and since $f(x) > 0$ for all $x \in [a, \infty)$, it follows that $\lim_{x \to \infty}f(x) = A$ exists and $A \geq 0$.
Next we can see that $f''(x) < 0$ so that $f'(x)$ is strictly decreasing for all $x$ and hence either $\lim_{x \to \infty}f'(x) = B$ or $f'(x) \to -\infty$ as $x \to \infty$. In the first case if $B$ exists then we must have $B \leq f'(a) < 0$ (because $f'(x) < f'(a)$ for all $x \in (a, \infty)$). By Mean Value theorem we have $f(x) - f(x/2) = (x/2)f'(c)$ for some $c$ satisfying $x/2 < c < x$. Taking limits as $x \to \infty$ we get a contradiction as LHS tends to $A - A = 0$ and RHS diverges to $-\infty$ because $f'(c)$ tends to a negative value. Same contradiction holds if $f'(x) \to -\infty$.
It follows that our assumption of existence of $a$ with $f'(a) < 0$ is wrong. Thus $f'(x) \geq 0$ for all $x \geq 0$.
Update: From the above proof it is easy to see that the result in question is true even if we assume $f(x) > K$ for some constant $K$. It is not necessary for $f(x)$ to be positive (like in the original question), it just needs to be bounded below.
