If $\frac{\sin\alpha}{\sin\beta} \le 1+\epsilon$, then $\frac{\alpha}{\beta} \le 1+\sqrt\epsilon\;\;$ (for acute $\alpha$ and $\beta$) Prove the following:

For $0 \le \alpha,\beta \le \frac{\pi}{2}$, if 
  $$\frac{\sin \alpha}{\sin \beta} \le 1+\epsilon$$
  then $$\frac{\alpha}{\beta} \le 1+\sqrt\epsilon$$

This is from this paper, page no 5.
 A: It can be stated in the following way: if $x,y\in[0,1]$ and $\frac{x}{y}\leq 1+\epsilon$, then $\frac{\arcsin(x)}{\arcsin(y)}\leq 1+\sqrt{\epsilon}$.
Or in the following way: if $z,w\in\mathbb{R}^-$ and $z-w\leq\log(1+\epsilon)$, then $\log\arcsin e^z-\log\arcsin e^w\leq \log(1+\sqrt{\epsilon})$. Or in the following way: for every $u>0$ and for every $z,w\in\mathbb{R}^-$ such that $z-w\leq u$, we have:
$$\log\arcsin e^{z}-\log\arcsin e^w \leq \log\left(1+\sqrt{e^u-1}\right).\tag{1}$$
Since $f(z)=\log\arcsin e^z$ is a convex function$^{(*)}$ on $\mathbb{R}^-$, for a fixed value of $z-w$ the maximum of the LHS of $(1)$ is attained when $z=0$. On the other hand the RHS of $(1)$ is an increasing function with respect to $u$, so the problem boils down to proving that:
$$\forall u>0,\qquad \frac{\pi}{2\arcsin e^{-u}}\leq 1+\sqrt{e^u-1} \tag{2}$$
or:
$$\forall t>0,\qquad \frac{\pi}{2\arcsin\frac{1}{t^2+1}}\leq 1+t \tag{3} $$
that, unfortunately, does not hold (the opposite inequality holds by the convexity of the LHS).
In order that grant the truth of your statement, the original $\frac{\pi}{2}$ has to be replaced by something smaller, or the original $\sqrt{\epsilon}$ has to be replaced by something like $\frac{8}{\pi}\sqrt{\epsilon}$, like the authors of the linked article do. The fixed inequality can be proved along the same lines as above.

$(*)$ Proof that $\log\arcsin e^x$ is a convex function on $\mathbb{R}^-$. It is enough to prove that
$$\frac{d}{dx}\log\arcsin e^x = \frac{e^x}{\sqrt{1-e^{2x}}\arcsin(e^x)}$$
is an increasing function on $\mathbb{R}^-$, or that
$$ \frac{u}{\sqrt{1-u^2}\arcsin(u)} $$
is an increasing function on $(0,1)$, or that
$$ \frac{\tan t}{t} $$
is an increasing function on $\left(0,\frac{\pi}{2}\right)$, but that easily follows from the convexity of the tangent function on such interval, or from the fact that the Taylor series of $\tan(x)$ in a neighbourhood of the origin has only non-negative coefficients, also because:
$$ \frac{\tan t}{t}=\sum_{n\geq 0}\frac{8}{(2n+1)^2\pi^2-4t^2}.$$
