Incircle of $A_1A_2A_3A_4$ is touching the sides $A_1A_2$, $A_2A_3$, $A_3A_4$, $A_4A_1$ at $K,L,M,N$ respectively. Let $A_1K=x$, $A_2L=y$, $A_3M=z$, $A_4N=t$. Finally lets say $A_1A_3=p$, $A_2A_4=q$. We have $2(x+y+z+t)=p_1$, $p+q=k_1$. Let $A_1A_3\cap A_2A_4=\{X\}$. We must have $\angle{A_1XA_2}\leq90^o$ or $\angle{A_2XA_3}\leq90^o$. WLOG let $\angle{A_1XA_2}\leq90^o.$ Let $Y$ be the reflection of $A_1$ on the midpoint of the segment $A_3A_4$. Then $A_1A_3YA_4$ is a parallelogram. $\Rightarrow A_4Y=A_1A_3=p, A_3Y=A_4A_1=x+t.$ Also $\angle{YA_4A_2}=\angle{A_3XA_2}=180^o-\angle{A_1X_A2}\geq90^o$. Lastly from triangle inequality yields $$x+y+z+t=(y+z)+(x+t)=A_2A_3+A_3Y\geq A_2Y.$$ Lets give a look onto the triangle $A_2A_4Y$. We have $$A_4Y=p, A_2A_4=q, A_2Y\leq x+y+z+t, \angle{A_2A_4Y}\geq 90^o.$$For they give a combined result of $$A_4Y^2+A_4A_2^2\leq A_2Y^2\Leftrightarrow p^2+q^2\leq (x+y+z+t)^2\Rightarrow \boxed{k_1^2}=(p+q)^2\leq2(p^2+q^2)\leq \frac{1}{2}.(2(x+y+z+t))^2=\boxed{\frac{1}{2}.p_1^2}\Rightarrow $$$$p_1^2\geq2k_1^2$$ Similarly we get $p_2^2\geq 2k_2^2$, so $$2(k_1^2+k_2^2)\geq (k_1+k_2)^2=p_1^2+p_2^2\geq 2k_1^2+2k_2^2$$ So ALL of our inequalities must hold the equality case. $(i)$ $\angle{A_1XA_2}=90^o$. $(ii)$ $A_2A_3+A_3Y= A_2Y.\Rightarrow$ $A_2,A_3,Y$ collinear. $\Rightarrow A_1A_4\parallel A_2A_3$ $(iii)$ $(p+q)^2=2(p^2+q^2).\Rightarrow$ $p=q$. $(iv)$ $2(k_1^2+k_2^2)= (k_1+k_2)^2\Rightarrow$ $k_1=k_2$. Note that similar equality cases for $B_1B_2B_3B_4$ also must hold. You can see that $(ii),(iii)$ shows that $A_1A_2A_3A_4$ is a isosceles trapezoid and $(i)$ yields $A_1A_2A_3A_4$ must be a square and $(iv)$ yields they are congruent.