Extended Sines of law!

since the traingle is an issoceles, it has angles $\left(A,\dfrac{\pi}{2}-\dfrac{A}{2},\dfrac{\pi}{2}-\dfrac{A}{2}\right)$ . from the constraints of the problem: $a=\dfrac{k_1+k_2}{2}, b=c=\sqrt{k_1k_2}$ . we use $\sin(B)=\sin\left(\dfrac{\pi}{2}-\dfrac{A}{2}\right) =\cos(A/2)$ and same for C, and $\sin(2B)=\sin(\pi-A)=\sin(A)$ to reduce the expression to $4^4 \dfrac{2\cos^{12} (A/2)}{(2\sin(A)\cos(A)+2\sin(A))^4}+1=\dfrac{2\cos^{12} (A/2)}{\sin^4(A)(\frac{\cos(A)+1}{2})^4}+1=\dfrac{2\cos^{12} (A/2)}{\sin^4(A)\cos^8(A/2)}+1\\ \dfrac{2\cos^4(A/2)}{(2\sin(A/2)\cos(A/2))^4}+1=\dfrac{1}{8 \sin^4(A/2)}+1$ various double and half angle identity was used above. by laws of cosine, for an isosceles triangle, we know $2b\sin(A/2)=a \to \sin(A/2)=\dfrac{a}{2b}=\dfrac{k_1+k_2}{4\sqrt{k_1k_2}}$ plugging this back in : $\dfrac{1}{8 \sin^4(A/2)}+1= 32\dfrac{k_1^2k_2^2}{(k_1+k_2)^4}+1= 32\dfrac{(k_1/k_2)^2}{(1+k_1/k_2)^4}+1$ we can minimize the function by minimizing $32\dfrac{r^2}{(1+r)^4}+1\to \dfrac{d}{dr} \dfrac{r^2}{(1+r)^4}=0\to \dfrac{2r}{(1+r)^4}-\dfrac{4r^2}{(1+r)^5}=0\to 2r(1+r)-4r^2=2r-2r^2=0\to r= 1,0$ its obvious that 0 is not the answer, hence using one we have the expression equal $\boxed{3}$ this is reached if $k_1=k_2$