You do understand this is likely impossible to do?
If there are ANY solutions which satisfy the given equality, then there are almost certainly infinitely many solutions. So finding ALL such solutions will take an infinitely long time to perform.
We can probably even show there are infinitely many solutions too. For example...
syms a b d s x y z
f = 1;
eqn = (a^2/x^2)-((a*(2*b))/x^2)+(2*(b^2)/y^2)+((1/x^2)-(2/y^2))*b^2+((2*(2*d)^2)/s^2)+(((y^2)^-1)-(4*x^2)^-1)*((2*z)^2) - f;
Now, we can search for one such solution. Does one exist? For example, suppose we set a=b=s=0.5. Now you want y < 2*x, so I'll pick values for x and t that satisfy that relationship. Say y = 0.5, and x = 0.3. Now look at what we get.
pretty(subs(eqn,[a,b,x,y,s],[.5 .5 .3 .5 .5]))
This represents an ellipse in the plane, with variables d and z. It has center (0,0).
fimplicit(subs(eqn,[a,b,x,y,s],[.5 .5 .3 .5 .5]),[0 1 0 1])
axis equal
grid on
xlabel 'd'
ylabel 'z'
So as long as d is a little smaller than 0.2, then there are infinitely many solutions, all living on that ellipse. We can even solve for z as a function of d.
z_d = solve(subs(eqn,[a,b,x,y,s],[.5 .5 .3 .5 .5]),z)
We will choose only the positive root there, so the second one found. It can be written as:
vpa(z_d(2))
And clearly that root will be real (for non-negative values of d) as long as d <= sqrt(1/32).
sqrt(1/32)
So no, you cannot find ALL solutions. There is NO optimization tool that will do so. I could have chosen many sets of the variables, picking random values for them from the domain of interest. How about if we pick
a0 = rand()
b0 = rand()
d0 = rand()*.2
z0 = rand()
s0 = rand()
eqnsubs = subs(eqn,[a,b,d,z,s],[a0,b0,d0,z0,s0])
fimplicit(eqnsubs,[0 1 0 1])
hold on
fimplicit(y == 2*x,[0 1 0 1],'r')
Here we see that anything below the red curve has y <= 2*x. So there will be infinitely many solutions, all of which lie along the blue curve for that set of random parameters.
I might also point out that I had to try multiple random values for a,b,d,s and z to find a set that had any solutions at all for x and y. There were many such sets of those parameters that failed to have solutions in the unit square, [0 1]X[0 1].
