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I'm new to Mathematica, so my question may be quite trivial, but I need help.

Let $F(x,y,a,b,c,d):\mathbb{R}^6\rightarrow \mathbb{R}$ be a function in 6 variables.

I need to find conditions for $x$ and $y$ such that, for $x$ and $y$ which respect these conditions, $F(x,y,a,b,c,d)=0$ is verified for every $a,b,c,d$ in a given range (i.e. $a_1\le a\le a_2$ and so on for $b,c,d$)

I've tried the command $Solve[F(x,y,a,b,c,d)==0,\{x,y\}]$ but then I don't know how to impose the constrains on $a,b,c,d$

How should I do it?

Thank you.

EDIT: I adapted Eric Towers' code :

F[x_,y_,a_,b_,c_,d_]:= Max[{a,b}] -2*((x*(a*x + (-c-d)*y) + y((c+d)x + by))/(x^2+y^2))

Assuming[{-1/2 < a< 1/2,a!=0, -1/2 < b < 1/2,b!=0,-1/2 < c < 1/2,c!=0,-1/2 < d < 1/2,d!=0},Simplify[Reduce[{F[x,y,a,b,c,d]>=0},{x,y},Reals]]]

But then in output I get

((x != 0 || y != 0) && (x == 0 || a <= 0 || Sqrt[(a x^2)/(a - 2 b)] <= y || Sqrt[(a x^2)/(a - 2 b)] + y <= 0) && b <= 0) || (b > 0 && ((Sqrt[((-2 a + b) x^2)/b] + y >= 0 && Sqrt[((-2 a + b) x^2)/b] >= y && 2 a < b && x != 0) || (2 a == b && y == 0 && x != 0) || (a == 2 b && x == 0 && y != 0) || (a > 2 b && (x != 0 || y != 0) && (x == 0 || y >= Sqrt[(a x^2)/(a - 2 b)] || Sqrt[(a x^2)/(a - 2 b)] + y <= 0))))

Which imposes conditions on a,b,c,d. But I would like solutions (x,y) for which F=0 is satisfied for every a,b,c,d (i.e. no condition on a,b,c,d). How should I fix it?

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  • $\begingroup$ Please show the actual Mathematica code that you are using, so that readers can provide more constructive feedback. $\endgroup$ – bbgodfrey May 25 '16 at 15:00
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No idea what your F looks like, so I made a simple one.

For your constraints, I'd use Assuming[] as an outer wrapper and apply Simplify[] to the results of Reduce[] (since you seem to be working with positive dimensional solution sets instead of sets of discrete points).

F[x_, y_, z_, u_, v_, w_] := x + y + Max[{z, u, v, w, 0}]

Assuming[{
    -1 <= z <= 0,
    -1 <= u <= 0,
    -1 <= v <= 0,
    -1 <= w <= 0
  },
  Simplify[
    Reduce[
      F[x, y, z, u, v, w] == 0,
      {x, y},
      Reals
    ]
  ]
]

(* x+y == 0 *)

EDIT: Now that we know the actual shape of your problem...

FullSimplify[
  Reduce[
    ForAll[{a, b, c, d}, And[
     -1/2 < a < 1/2, -1/2 < b < 1/2, -1/2 < c < 1/2, -1/2 < d < 1/2
      ],
    F[x, y, z, u, v, w] >= 0
    ],
    {x, y},
    Reals
]]

(*  (z <= 0 && (((-Sqrt[((x^2 (u - 2 z))/u)] <= y <= Sqrt[(x^2 (u - 2 z))/ u] || u <= 0) && x != 0) || (u <= 0 && y != 0))) || 
    (z > 0 && ((2 u < z && (x != 0 || y != 0) && (x == 0 || y >= Sqrt[(x^2 z)/(-2 u + z)] || y + Sqrt[(x^2 z)/(-2 u + z)] <= 0)) || (2 u == z && x == 0 && y != 0) || (u == 2 z && y == 0 && x != 0) || (y + Sqrt[(x^2 (u - 2 z))/u] >= 0 && Sqrt[(x^2 (u - 2 z))/u] >= y && x != 0 && u > 2 z)))  *)
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