$f+f'+f''\geq0$,Prove the $f$ has a lower bound

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

$$f(t)=e^{t/2}g(\tfrac{\sqrt3}2t)\geq e^{a/2}C.$$