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.
 A: As pointed out in the comments the inequality:
$$\large \frac{1}{AM \cdot BN} + \frac{1}{BN \cdot CP} + \frac{1}{CP \cdot AM} \le \frac{4}{3(R - OI)}$$
Is not homogeneous and therefore cannot be correct. Take any triangle and any point and even if the given inequality is satisfied fot this configuration then after scaling it by $a$ for sufficiently small $a$ it will stop being correct. But somehow you have managed to convert it to homogeneous inequality:
$$\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI} \le \frac{2}{R - OI}$$
Which still seems not to be true. And even if it was you obtained it by means of this wrong inequality (no surprise since homogeneous and non-homogeneous inequalities can't be equivalent):
$$\left(\frac{1}{AM} + \frac{1}{BN} + \frac{1}{CP}\right)^2 \le \left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)\left(\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI}\right)$$
Which is a flawed application of CS inequality. The correct application is:
$$\left(\frac{1}{\sqrt{AM}} + \frac{1}{\sqrt{BN}} + \frac{1}{\sqrt{CP}}\right)^2 \le \left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)\left(\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI}\right)$$
A: It's wrong.
Indeed, let $\Delta A'B'C'\sim\Delta ABC$ such that $\frac{A'B'}{AB}=\epsilon.$
Thus, if your inequality is true, so we have $$\sum_{cyc}\frac{1}{A'M'\cdot B'N'}\leq\frac{4}{3(R'-O'I')}$$ or
$$\frac{1}{\epsilon}\sum_{cyc}\frac{1}{AM\cdot BN}\leq\frac{4}{3(R-OI)},$$ which is wrong for $\epsilon\rightarrow0^+$.
A: 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
$$\left\{ \begin{align} AI \ge |AO - OI|\\ BI \ge |BO - OI|\\ CI \ge |CO - OI|\\ \end{align} \right. = |R - OI|$$
