<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] 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:
$$\left(\frac{1}{\sqrt{AM}} + \frac{1}{\sqrt{BN}} + \frac{1}{\sqrt{CP}}\right)^2 \le \left(\frac{AI}{AM} + \frac{BI}{BN} + \frac{CI}{CP}\right)\left(\frac{1}{AI} + \frac{1}{BI} + \frac{1}{CI}\right)$$
</s>