If the answer to this question is in affirmative, then this would yield a good answer to [this question](https://mathoverflow.net/questions/342763). 


Let $f\colon \mathbb C\to\mathbb C$ be an entire function whose values on the real line are real and bounded from below by
$$
 f(x)\geq \sqrt{\frac{\sinh x}{x}}, \qquad 0\neq x\in\mathbb R.
$$
**[EDIT]** Also, we have $|f(x+iy)|\leq C\exp(|x|/2)$ and, moreover, $f(x+iy)\to0$ as $y\to\infty$ for each separate $x$.


Is it true that $f(0)\geq 2/\sqrt\pi$?


**Remarks.** **1.** If so, this estimate is sharp and achieved, e.g., on the analytic continuation of
$$
 f(x)=e^{x/2}\sqrt{\frac8{\pi x}}\int\_0^{\sqrt {x/2}}\exp(-t^2) 
 \,\mathrm{d}t
 =\frac{e^{x/2}}{\sqrt{\pi x}}\int\_0^{x}\frac{e^{-s/2}}{\sqrt s}
 \,\mathrm{d}s, 
 \qquad x>0.
$$
(The fact that this function satisfies the inequality follows from the connection to the question mentioned above.)


**2.** It is worth nothing to mention that one may assume $f$ to be even. 


**3.** Clearly, $\sqrt{\sinh z/z}$ is not an entire function, as $\sinh z$ has many simple zeroes.


**4.** Only the first additional constraint $|f(x+iy)|\leq C\exp(|x|/2)$ does not exclude the counterexample $\cosh(z/2)$; thanks to Michael Engelhardt who noticed that I need some extra conditions.



Entire function not less than $\sqrt{\sinh x/x}$ on the real line
The answer is no. Let $f$ be the function described under Remark 1 of the OP. By continuity, there exists a $\delta >0$ such that $f(x) - \sqrt{(\sinh x)/x} > 0.05$ on the real axis for $|x|\leq \delta $ (note that, at $x=0$, the aforementioned difference is $2/\sqrt{\pi } -1 = 0.128$).


Consider furthermore the function
$$
g(z)=\epsilon \left[\frac{2\sinh (z/2)}{z}-(1+\kappa )\frac{\pi^{2} \cosh (z/2)}{z^2+\pi^{2} } \right]
$$
$g$ satisfies the required constraints $|g(x+iy)| \leq C\exp (|x|/2)$ and $g(x+iy)\rightarrow 0$ as $y\rightarrow \infty$ for each separate $x$. On the real axis, $g(x)\geq 0$ for $\kappa =0$; we can adjust to a $\kappa > 0$ such that $g(x)\geq 0$ for $|x|\geq \delta $ but $g(x)<0$ for $|x|<\delta $. We can furthermore adjust $\epsilon $ such that $|g(x)|<0.05$ for $|x|< \delta $. With such a choice of $\kappa $ and $\epsilon $, the function $f+g$ is a counterexample to the OP's question, $f(0)+g(0) < 2/\sqrt{\pi } $ while satisfying all the given constraints.


