Minimizing Inequality

For this, we make use of a simplified version of Cauchy Schwartz inequality, which states; $\frac{a^2}{b}+\frac{c^2}{d}+\frac{e^2}{f}≥\frac{(a+c+e)^2}{b+d+f}$ Therefore , moving to the question again, (Here, I will ignore the outer square this moment and use it later)

So, by applying Cauchy Schwartz inequality, we get, $\frac{1}{1+2ab-c^2}+\frac{1}{1+2bc-a^2}+\frac{1}{1+2ca-b^2}≥\frac{(1+1+1)^2}{1+1+1-(a^2+b^2+c^2)+2(ab+bc+ca)}$ Which gives ; $≥\frac{9}{2+2(ab+bc+ca)}..............................(1)$

Now by simple $AM-GM$ inequality, we get, $a^2+b^2+c^2≥ab+bc+ca$

Now, in order to minimize $(1)$ , we must consider the equality case and should take the denominator to be maximum, So, $ab+bc+ca$ is maximum at $1$ , since $a^2+b^2+c^2=1$

Therefore, putting the above results in $(1)$ , we get the minimum as, $Minimum=\frac{9}{2+2}=\frac{9}{4}$

Therefore, according to the question, we get, Minimum of

$(\frac{1}{1+2ab-c^2}+\frac{1}{1+2bc-a^2}+\frac{1}{1+2ca-b^2})^2=(\frac{9}{4})^2=\frac{81}{16}=\frac{p}{q}$ Therefore, $p+q=81+16=97$

We want to find the minimum value of $\left(\sum_{cyc}\dfrac{1}{1+2ab-c^2}\right)^2$

First, homogenize by multiplying by $(a^2+b^2+c^2)^2$ : $(a^2+b^2+c^2)^2\left(\sum_{cyc}\dfrac{1}{1+2ab-c^2}\right)^2$

Juggle some powers: $\left((a^2+b^2+c^2)\sum_{cyc}\left(\dfrac{1}{1+2ab-c^2}\right)\right)^2$

We apply Cauchy on the inside to get $\left((a^2+b^2+c^2)\sum_{cyc}\left(\dfrac{1}{1+2ab-c^2}\right)\right)^2\ge \left(\sum_{cyc}\dfrac{a}{\sqrt{1+2bc-a^2}}\right)^4$

Substituting $1-a^2=b^2+c^2$ to each of the fractions: $\left(\sum_{cyc}\dfrac{a}{\sqrt{1+2bc-a^2}}\right)^4= \left(\sum_{cyc}\dfrac{a}{\sqrt{b^2+2bc+c^2}}\right)^4$

Thankfully, this gets rid of the square roots: $\left(\sum_{cyc}\dfrac{a}{\sqrt{b^2+2bc+c^2}}\right)^4= \left(\sum_{cyc}\dfrac{a}{b+c}\right)^4$

However, $\sum_{cyc}\dfrac{a}{b+c}\ge \dfrac{3}{2}$ is true because that is the exact statement of Nesbitt's Inequality. Thus, $\left(\sum_{cyc}\dfrac{a}{b+c}\right)^4\ge \left(\dfrac{3}{2}\right)^4=\dfrac{81}{16}$

Thus, our answer is $81+16=\boxed{97}$

Equality case is when $a=b=c=\dfrac{\sqrt{3}}{3}$ .