Here is an explicit solution found computationally (and then beautified) using the approach referenced by Dima.

Let $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt),$$ $$g(x,y,z,t) := (x-y) (z-t) (xy(z-t) + (x-y)zt).$$ Then the discriminant in question equals $$2a^{16}(7s_1 + s_2),$$ where $$s_1:= f(a_1^2,a_2^2,a_3^2,a_4^2)^2 + f(a_1^2,a_3^2,a_2^2,a_4^2)^2 + f(a_2^2,a_3^2,a_1^2,a_4^2)^2$$ and $$s_2:= g(a_1^2,a_2^2,a_3^2,a_4^2)^2 + g(a_1^2,a_3^2,a_2^2,a_4^2)^2 + g(a_2^2,a_3^2,a_1^2,a_4^2)^2 + g(a_3^2,a_2^2,a_1^2,a_4^2)^2 + g(a_1^2,a_1^2,a_3^2,a_4^2)^2 + g(a_3^2,a_1^2,a_2^2,a_4^2)^2.$$ (Btw, it can be verified that both $s_1$ and $s_2$ represent symmetric polynomials in $a_1^2,a_2^2,a_3^2,a_4^2$.)

Now, we see that the discriminant as the sum of squares can be zero only when all these squares are zero. Since $a_i$ are pairwise distinct, we can cancel the first two (linear) factors in $f,g$ and focus on third factors being zero. However, in the ideal generated by these factors, there is a polynomial (I checked the first one in the Grobner basis) that is nonzero for distinct nonzero $a_i$, meaning that all squares cannot be zero at the same time.

So, the discriminant is strictly positive.

Here is an explicit solution found computationally (and then beautified) using the approach referenced by Dima.

Let $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt),$$ $$g(x,y,z,t) := (x-y) (z-t) (xy(z-t) + (x-y)zt).$$ Then the discriminant in question equals $$2a^{-8}(7s_1 + s_2),$$ where $$s_1:= f(a_1^2,a_2^2,a_3^2,a_4^2)^2 + f(a_1^2,a_3^2,a_2^2,a_4^2)^2 + f(a_2^2,a_3^2,a_1^2,a_4^2)^2$$ and $$s_2:= g(a_1^2,a_2^2,a_3^2,a_4^2)^2 + g(a_1^2,a_3^2,a_2^2,a_4^2)^2 + g(a_2^2,a_3^2,a_1^2,a_4^2)^2 + g(a_3^2,a_2^2,a_1^2,a_4^2)^2 + g(a_1^2,a_1^2,a_3^2,a_4^2)^2 + g(a_3^2,a_1^2,a_2^2,a_4^2)^2.$$ (Btw, it can be verified that both $s_1$ and $s_2$ represent symmetric polynomials in $a_1^2,a_2^2,a_3^2,a_4^2$.)

Now, we see that the discriminant as the sum of squares can be zero only when all these squares are zero. Since $a_i$ are pairwise distinct, we can cancel the first two (linear) factors in $f,g$ and focus on third factors being zero. However, in the ideal generated by these factors, there is a polynomial (I checked the first one in the Grobner basis) that is nonzero for distinct nonzero $a_i$, meaning that all squares cannot be zero at the same time.

So, the discriminant is strictly positive.

PS. In fact, $s_2$ alone cannot be zero for pairwise distinct $a_i$.

Here is an explicit solution found computationally (and then beautified) using the approach referenced by Dima.

Let $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt),$$ $$g(x,y,z,t) := (x-y) (z-t) (xy(z-t) + (x-y)zt).$$ Then the discriminant in question equals $$2a^{16}(7s_1 + s_2),$$ where $$s_1:= f(a_1^2,a_2^2,a_3^2,a_4^2)^2 + f(a_1^2,a_3^2,a_2^2,a_4^2)^2 + f(a_2^2,a_3^2,a_1^2,a_4^2)^2$$ and $$s_2:= g(a_1^2,a_2^2,a_3^2,a_4^2)^2 + g(a_1^2,a_3^2,a_2^2,a_4^2)^2 + g(a_2^2,a_3^2,a_1^2,a_4^2)^2 + g(a_3^2,a_2^2,a_1^2,a_4^2)^2 + g(a_1^2,a_1^2,a_3^2,a_4^2)^2 + g(a_3^2,a_1^2,a_2^2,a_4^2)^2.$$ (Btw, it can be verified that both $s_1$ and $s_2$ represent symmetric polynomials in $a_1^2,a_2^2,a_3^2,a_4^2$.)

Now, we see that the discriminant as the sum of squares can be zero only when all squares are zero. Since $a_i$ are pairwise distinct, we can cancel the first two (linear) factors in $f,g$ and focus on third factors being zero. However, in the ideal generated by these factors, there is a polynomial (I checked the first one in the Grobner basis) that is nonzero for distinct nonzero $a_i$, meaning that all squares cannot be zero at the same time.

So, the discriminant is strictly positive.

Here is an explicit solution found computationally (and then beautified) using the approach referenced by Dima.

Let $$f(x,y,z,t) := (x-y) (z-t) (xy(z+t) - (x+y)zt),$$ $$g(x,y,z,t) := (x-y) (z-t) (xy(z-t) + (x-y)zt).$$ Then the discriminant in question equals $$2a^{16}(7s_1 + s_2),$$ where $$s_1:= f(a_1^2,a_2^2,a_3^2,a_4^2)^2 + f(a_1^2,a_3^2,a_2^2,a_4^2)^2 + f(a_2^2,a_3^2,a_1^2,a_4^2)^2$$ and $$s_2:= g(a_1^2,a_2^2,a_3^2,a_4^2)^2 + g(a_1^2,a_3^2,a_2^2,a_4^2)^2 + g(a_2^2,a_3^2,a_1^2,a_4^2)^2 + g(a_3^2,a_2^2,a_1^2,a_4^2)^2 + g(a_1^2,a_1^2,a_3^2,a_4^2)^2 + g(a_3^2,a_1^2,a_2^2,a_4^2)^2.$$ (Btw, it can be verified that both $s_1$ and $s_2$ represent symmetric polynomials in $a_1^2,a_2^2,a_3^2,a_4^2$.)

Now, we see that the discriminant as the sum of squares can be zero only when all these squares are zero. Since $a_i$ are pairwise distinct, we can cancel the first two (linear) factors in $f,g$ and focus on third factors being zero. However, in the ideal generated by these factors, there is a polynomial (I checked the first one in the Grobner basis) that is nonzero for distinct nonzero $a_i$, meaning that all squares cannot be zero at the same time.

So, the discriminant is strictly positive.