$f+f'+f''\geq0$,Prove the $f$ has a lower bound Let $f\in C^2(a,b)$ such that $f+f'+f''\geq0$ 
Prove that $f$ has a lower bound.
$My\quad Attempt$
$1.\quad$Suppose that $f$ has no lower bound at x=b,so there is a sequence$\{x_n\}$,which converges to b($\lim_{n\to\infty}{x_n}=b$),and $f'(x_n)<0,f(x_n)<-n$
$2.\quad\forall k\in \mathbb N,\exists N\in\mathbb N,n>N,x_n>x_k$,then prove that $$\int_{\{x|x\in(x_k,x_n)\land f(x)>0\}}{}f^2(x)dx\leq C$$
$3.\quad$Prove: $$0\leq\int_{x_k}^{x_n}{(f+f'+f'')}dx\leq0+f(x_k)-f(x_n)+C\to-\infty$$
$\quad\quad\quad\quad\quad\quad\quad$This contradicts the problem
So that's my idea，but I can't do it from step 2.And my idea might be wrong.
Edit in 2019/2/16
I solve the question that if $q=0$.There is some Chinese characters in my answer.I hope it doesn't bother you

 A: Lemma. Given $a<b,$ any $g\in C^2(a,b)$ satisfying $g''+g\geq 0$ is bounded below.
Proof. We can apply (basically) the Picone identity, i.e. the derivative of a Wronskian. Specifically, for $s,t$ satisfying $\max(a,b-\pi)<t<s<b$ define $h_s(t)=g(t)\cos(t-s)-g'(t)\sin(t-s).$ Then
$$h'_s(t)=(g(t)\cos(t-s)-g'(t)\sin(t-s))'=-(g(t)+g''(t))\sin(t-s)\geq 0$$
because $\sin(t-s)<0.$ So
$$g(s)=h_s(s)\geq h_s(t)\geq -|g(t)|-|g'(t)|.$$
Picking any $\max(a,b-\pi)<t<b,$ this shows that $g$ is bounded below on $(t,b).$
The function $G\in C^2(a,b)$ defined by $G(x)=g(a+b-x)$ satisfies
$$G''(t)+G(t)=g''(a+b-t)+g(a+b-t)\geq 0.$$
So the same argument shows that $G$ is bounded below on $(t,b).$ This means $g$ is bounded below on $(a,a+b-t),$ and we previously showed that $g$ is bounded below on $(t,b).$ By continuity $g$ is bounded below on the whole interval $(a,b).$
Corollary. Given $a<b,$ any $f\in C^2(a,b)$ satisfying $f''+f'+f\geq 0$ is bounded below.
Proof. The transformation $g(t)=e^{t/\sqrt 3}f(\tfrac{2}{\sqrt 3}t)$ gives
$$g''(t)+g(t)=e^{t/\sqrt 3}(\tfrac{4}{3}f''(\tfrac{2}{\sqrt 3}t)+\tfrac{4}{3}f'(\tfrac{2}{\sqrt 3}t)+\tfrac{4}{3}f(\tfrac{2}{\sqrt 3}t))\geq 0$$
for $a<\tfrac{2}{\sqrt 3}t<b.$ The function $g$ therefore satisfies $g''+g\geq 0$ on $(\tfrac{\sqrt 3}{2}a,\tfrac{\sqrt 3}{2}b).$ By the lemma, there exists a real number $C$ such that $g(t)\geq C$ for all $t\in (\tfrac{\sqrt 3}{2}a,\tfrac{\sqrt 3}{2}b).$ Therefore
$$f(t)=e^{t/2}g(\tfrac{\sqrt3}2t)\geq e^{a/2}C.$$
A: Proof:  Let  $F(x) = e^{\frac{x}{2}}f(x)$,then
​        $$\begin{align}\frac{3}{4}F(x)+F''(x) &=\frac{3}{4} e^{\frac{x}{2}}f(x)+e^{\frac{x}{2}}(\frac{1}{4}f(x)+f'(x)+f''(x))\\ &=e^{\frac{x}{2}}(f(x)+f'(x)+f''(x))\\ &\geq0 \end{align}$$
​       (1) $F(x)<0,\forall x\in(a,b).$
​          $F''(x)\geq -F(x)\geq 0$, so $F(x)$ is convex function. Picking any $x_0\in(a,b)$,we have
​                   $$F(x)\geq F(x_0)+F'(x_0)(x-x_0),\quad\quad\quad\quad\forall x\in(a,b).$$
​       (2) $\exists x_0\in(a,b),F(x_0)\geq0$
​          We suppose that there's  $n$ zero points in $(a,b)$ at most, and $2\leq n<\infty$.
​           For $0\leq n\leq 1$ , we assume that $F(x)<0 ,x\in(a,x_0),F(x)\geq 0,x\in[x_0,b),$
​           In $(a,x_0)$, we know that $F(x)$  has a lower bound from 1 
​           In $[x_0,b)$, $F(x) $ has a lower bound obviously.
So we get $n\geq2$,  ​$\exists \alpha,\beta\in(a,b),F(\alpha)=F(\beta)=0$, and
​                         $$F(x)\geq 0, \quad \quad\quad\quad\quad\quad x\in(\alpha,\beta).$$
​          Let $\gamma\in(\alpha,\beta)$ is absolute maximum point of $F(x)$ in $[\alpha,\beta]$, then for $t>\gamma,F''(t)\leq0$.
​         Then $\exists s\in[\gamma,t],F'(s)=0$,
​                       $$F'(x)<0,\quad \quad\quad\quad\quad\quad x\in(s,t)$$
​          So $|F'(x)|^2+\frac{3}{4}|F(x)|^2$ is monotone decreasing in $[s,t]$ ,then
​                   $$\begin{align}F'(x)&\geq-\sqrt{|F'(s)|^2+\frac{3}{4}|F(s)|^2-\frac{3}{4}|F(x)|^2}\\ &=-\frac{3}{4}\sqrt{|F(s)|^2-|F(x)|^2}\\ &\geq-\frac{3}{4}\sqrt{|F(\gamma)|^2-|F(x)|^2},\quad x\in(s,t) \end{align}$$
​           We get,
​               $$F'(x)\geq-\frac{3}{4}\sqrt{|F(\gamma)|^2-|F(x)|^2},\quad x\in[\gamma,\beta]$$
​           that is, $arcsin\frac{F'(x)}{F'(\gamma)}+\frac{3}{4}x $  is monotone increasing in $[\gamma,\beta]$, then 
$$
\frac{3\beta}{4}=arcsin\frac{F'(\beta)}{F'(\gamma)}+\frac{3}{4}x\geq arcsin\frac{F'(\gamma)}{F'(\gamma)}+\frac{3}{4}x=\frac{\pi}{2}+\frac{3\gamma}{4}​
$$
​           which implies $\beta-\alpha\geq\beta-\gamma\geq\frac{2\pi}{3}$, we get  $2\leq n<\infty$
​               If $F(x)$ doesn't have a lower bound in $[x_0,b)$, $\exists x_1>x_0,F'(x_1)=0$, and $F(x)<0,x\in[x_1,b)$,
​           $$F(x)\geq F'(x_1)(x-x_1),\quad \quad\forall x\in[x_1,b).$$
​           We get contradictions from the above formula.So $F(x)$ has a lower bound in$[x_0,b)$.
​           We can prove that $F(x)$ has a lower bound in$(a,x_0]$ by the same way.
​               Obviously, $f(x)$ has a lower pound in $(a,b)$.
