Prove that $\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}<\frac{2}{R'-OO'}$ Let $\Delta ABC$ is acute triangle has $AB>AC$ and $O$ is incircle of $\Delta ABC$ with radius $R$, $O'$ is circumcircle of $\Delta ABC$ with radius $R'$. $OA\cap BC=A';OB\cap AC=B';OC\cap AB=C'$ .Prove that $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}<\frac{2}{R'-OO'}$$


I proved that $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}=\frac{1}{\frac{2bc}{b+c}\cdot \cos \frac{A}{2}}+\frac{1}{\frac{2ac}{a+c}\cdot \cos \frac{B}{2}}+\frac{1}{\frac{2ab}{b+a}\cdot \cos \frac{C}{2}}$$
And by Euler's formula $OO'=\sqrt{R'^2-2R\cdot R'}$
$$\Rightarrow R'-OO'=\frac{abc}{4S}-\sqrt{\left(\frac{abc}{4S}\right)^2-2\cdot \frac{abc}{4S}\cdot \frac{S}{\frac{a+b+c}{2}}}$$
$$=\frac{abc}{4S}-\frac{a^4bc+abc^4+ab^4c+3a^2b^2c^2-a^3bc^2-a^3b^2c-a^2bc^3-a^2b^3c-ab^2c^3-ab^3c^2}{\left(a+b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)}$$
I tried to change this problem to an inequality with three variables $a,b,c$ but i think it's harder than before.

I shall use standard notation in my proof, that is, $R, r, s$ and $\Delta$ are the circumradius, inradius, semi-perimeter and area of the triangle, $AA'$ is the length of $A-\text{internal angle bisector}$ and other similarly.
$$rR(\sin A+\sin B+\sin C)=rs=\Delta=\frac{1}{2}ab\sin C=2R^2 \sin{A}\sin{B}\sin{C}$$$$\implies \frac{\sin A+\sin B+\sin C}{2R \sin{A}\sin{B}\sin{C}}=\frac{1}{r}$$
For completeness, I will include the derivation of $AA'$, which you have already done. Applying sine rule on triangle $AA'C$, we have, $$\frac{AA'}{\sin C}=\frac{b}{\sin{(B+A/2)}}\implies AA'=\frac{2R\sin B\sin C}{\cos\frac{B-C}{2}}\geq2R\sin B\sin C$$
$$\therefore \frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{\sin A+\sin B+\sin C}{2R \sin{A}\sin{B}\sin{C}}=\frac{1}{r}$$
Now, by Euler's formula, $OI=\sqrt{R^2-2Rr}$.
$$\therefore \frac{2}{R-OI}=\frac{2}{R-\sqrt{R^2-2Rr}}=\frac{R+\sqrt{R^2-2Rr}}{Rr}\geq\frac{1}{r}$$
Therefore, $\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{2}{R-OI}$, with equality iff the triangle is equilateral.
$\blacksquare$
Here is another proof which does not use trigonometry and is probably more enlightening. Let $h_a, h_b, h_c$ be the respective altitudes of $\triangle ABC$. Then, $AA'\geq h_a$ etc. So, $$\frac{1}{AA'}+\frac{1}{BB'}+\frac{1}{CC'}\leq\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{\Delta}\bigg(\frac{\Delta}{h_a}+\frac{\Delta}{h_b}+\frac{\Delta}{h_c}\bigg)=\frac{s}{\Delta}=\frac{1}{r}$$ The rest of the proof runs along the same lines as the last one. As can easily be seen from this proof, the result is nothing special about angle bisectors, that is, for any three cevians $l_a$,$l_b$ and $l_c$ from points $A$, $B$ and $C$, $$\frac{1}{l_a}+\frac{1}{l_b}+\frac{1}{l_c}\leq\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}=\frac{1}{r}\leq\frac{2}{R-OI}$$
