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Does every triangle satisfy $\frac{1}{a-b+\pi R} + \frac{1}{b-c+\pi R} - \frac{1}{c-a+\pi R} < \frac{1}{R}$?

Experimental data show that in any triangle with sides $(a,b,c)$ and circumradius $R$, if $x > 2$ then,

$$\frac{1}{x} - \frac{4}{x^2 - 4} < \frac{R}{a-b+Rx} + \frac{R}{b-c+Rx} - \frac{R}{c-a+Rx} < \frac{1}{x} + \frac{4}{x^2 - 4} \tag 1$$

Can this be proved or disproved?

Corollary: Assuming the above upper bound is true, at $x = \pi$, the upper bound is $0.999787$ which is remarkably close to $1$ and thus we have

$$\frac{1}{a-b+\pi R} + \frac{1}{b-c+\pi R} - \frac{1}{c-a+\pi R}< \frac{1}{R} \tag 2$$

Related question.


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