<s>[INST] Prove that $\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$. 

$(O, R)$ is the circumscribed circle of $\triangle ABC$. $I \in \triangle ABC$. $AI$, $BI$ and $CI$ intersects $AB$, $BC$ and $CA$ respectively at $M$, $N$ and $P$. Prove that $$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)^2}$$

I have provided my own solution and I would be greatly appreciated if there are any other solutions, perhaps one involving trigonometry. I deeply apologise for the misunderstanding.
 [/INST] Firstly,
$$\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP} = \frac{A_{CIB}}{A_{CAB}} + \frac{A_{AIC}}{A_{ABC}} + \frac{A_{BIA}}{A_{BCA}} = 1$$
where $A_m$ denotes the area of shape $m$.
$$\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP} = 3 - \left(\frac{IM}{AM} + \frac{IN}{BN} + \frac{IP}{CP}\right) = 2$$
We have that $$\frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{1}{3}\left(\frac{1}{AM} + \frac{1}{BN} + \frac{1}{CP}\right)^2$$
$$\le \frac{1}{3}\left[\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2\right]\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right)$$
(according to the Cauchy-Schwarz inequality)
However 
$$\left(\frac{AI}{AM}\right)^2 + \left(\frac{BI}{BN}\right)^2 + \left(\frac{CI}{CP}\right)^2 \ge \dfrac{1}{3}\left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)^2 = \frac{4}{3}$$
Now I just need to prove that
$$\frac{1}{3}\left(\frac{1}{AI^2} + \frac{1}{BI^2} + \frac{1}{CI^2}\right) \le \frac{1}{(R - OI)^2}$$
which can be easily proven because
$$\left\{ \begin{align} AI \ge |AO - OI|\\ BI \ge |BO - OI|\\ CI \ge |CO - OI|\\ \end{align} \right. = |R - OI|$$
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