Distances from Morley triangle to edges of the original triangle

Let $a,b,c$ be the sides of the triangle and, for simplicity, $A=3\alpha, B=3\beta, C=3\gamma$ be the angles, and hence $\alpha+\beta+\gamma=60^\circ$.

Regarding the triangle $\triangle AEC$ we have (recall the law of sines) $$\frac{AE}{\sin\gamma}=\frac{b}{\sin(180^\circ-\alpha-\gamma)}\iff AE=\frac{b\cdot \sin \gamma}{\sin(\alpha+\gamma)}$$ Consider now the right triangle $\triangle EHA$, where $$\sin\alpha=\frac{EH}{AE}\iff EH=AE\cdot \sin\alpha=\frac{b\cdot\sin\alpha\cdot\sin\gamma}{\sin(\alpha+\gamma)}$$ Now, have a looke here, where they prove (using the same notation) that $m:=EF=8R\sin\alpha\sin\beta\sin\gamma$. Thus $$\frac{m}{EH}=\frac{8R\sin\beta\cdot\sin(\alpha+\gamma)}{b}=\frac{8R\sin\beta\cdot \sin(60^\circ-\beta)}{b}=\frac{8R\cdot\sin\beta\cdot\cos(\beta+30^\circ)}{b}$$ The expansion of $\cos(x+y)=\cos x\cos y-\sin x\sin y$, yields $$\frac{m}{EH}=\frac{8R\sin\beta\cdot\left(\frac{\sqrt{3}}2\cos \beta-\frac12\sin\beta\right)}{b}=\frac{2R\cdot\left(\sqrt{3}\sin(2\beta)-2\sin^2\beta\right)}b$$ Finally, in virtue of the law of Sines, $b=2R\cdot \sin(3\beta)$. Thus $$\frac{m}{EH}=\frac{\sqrt{3}\sin(2\beta)-2\sin^2\beta}{\sin(3\beta)}:=f(\beta)$$ Considering that $60^\circ\geqslant\beta\geqslant0^\circ$, you can prove your inequality using calculus.

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EDIT: The acute observation @John Bentin made yields $$f(\beta)=\frac1{\cos\left(\beta-30^\circ\right)}$$ Since we are dealing with the cosine function, we know $\cos(\beta-30^\circ)$ will atain its maximum at $\beta=30^\circ$ which, thus is the minimum of $f$. Furthermore, since $0\leqslant\beta\leqslant 60^\circ$, the cosine function will atain ist minimum at the borders, i.e. at $\beta=0^\circ$ or $\beta=60^\circ$. This leads to the conclusion , that the minima and maxima are resprecitvely $\displaystyle \frac1{\cos 0^\circ}=1$ and $\displaystyle \frac1{\cos30^\circ}=\frac2{\sqrt{3}}$.