Prove that if $\forall x \in \mathbb{R} f''(x) \geq 0$ and $\lim\limits_{x \to \infty} f(x)=0$ then $\forall x \in \mathbb{R} f(x) \geq 0$ As stated in the title, I need help proving the following:
If $\forall x \in \mathbb{R} f''(x) \geq 0$ and $\lim\limits_{x \to \infty} f(x)=0$ then $\forall x \in \mathbb{R} f(x) \geq 0$
I have absolutely no idea as to how to even approach this. I know that $f'$ must be monotone, but I don't see how it helps me.

Any help will be appreciated, thanks in advance!
 A: By convexity we have $$f((x+y)/2)\le {1\over 2}[f(x)+f(y)]$$ Fix $x$ and let  $y\to \infty$. Then $f(x)\ge 0.$
A: Hint (almost solution).
$f''(x) \ge 0 \Longrightarrow$ $f'(x)$ is nondecreasing. Hence there are two variants:

*

*$f'(x) \le 0$ for all $x$


*$f'(x) > 0$ for $x > x_0$
Consider case $1)$. In this case $f(x)$ is nonincreasing and $f(x) \to 0$. It's easy to see that $f(x) \ge 0$ (use proof by Contradiction)
Consider case $2)$ and $x > x_0$. In this case $f(x)$ is nondecreasing and $f(x) \to 0$. Suppose that $f(a) < 0$ at some point $a$. We know that $f(x) \to 0$ as $x \to \infty$ and $f$ is continious hence there exists $b > a$ such that $f(b) = f(a)/2$.
Now consider three points: $ A =(a, f(a)), B = (b, f(b))$ and $C = (x, f(x))$ for some $x > b$. It's not hard to see (use a picture) that $B$ lies below $AC$ if $x$ is big enoungh. But $f''(x) \ge 0$ hence $f$ in concave. We got a contradiction.
A: By the convexity of $f$ the slope of the chord joining $[x,y]$ is atmost that of $[y,z]$ for $ x \leq y \leq z$, and taking $ z \to \infty$ gives $f(y) \leq f(x)$, which shows $f$ is decreasing, so that $f \geq 0$.
A: $f'$ is increasing because $f''\ge 0.$ If $f'(x_0)>0$ for some $x_0$ then for $x>x_0$ we would have
$f(x)=f(x_0)+\int_{x_0}^xf'(t)dt\ge f(x_0)+\int_{x_0}^xf'(x_0)dt=f(x_0)+(x-x_0)f'(x_0),$
which would imply $f(x)\to\infty$ as $x\to\infty,$ contrary to $\lim_{x\to\infty}f(x)=0.$
Therefore $f'(x)\le 0$ for all $x.$
So $f$ is decreasing. If $f(x_1)<0$ for some $x_1$ then for all $x>x_1$ we would have $f(x)\le f(x_1),$  contrary to $\lim_{x\to\infty}f(x)=0.$
Example: $f(x)=e^{-x}.$
A: Suppose $\ \exists\ a\in\mathbb{R}\ $ such that $\ f(a) <0.\ $ Since $\ \displaystyle\lim_{x\to\infty} f(x) = 0,\ \exists\ b > a\ $ such that $\ f(b) > f(a).\ $ But then due to convexity of $\ f,\ $ and the fact that $\ b > a\ $ and $\ f(b) > f(a)\ $ we can show that there is a point $\ c > b\ $ such that $\ f(x) \geq 1\ \forall\ x \geq c,\ $ contradicting $\ \displaystyle\lim_{x\to\infty} f(x) = 0.\ $ A contradiction has arisen, and so we reject our only further assumption: that $\ \exists\ a\in\mathbb{R}\ $ such that $\ f(a) <0.\ $
