Verify an inequality of many parameters
Making use of the valuable comments by flinty, I obtain
n = 2000; con = (k + μ)*(γ1 + μ) - β1*n*
k*(1 - q)*(γ2 + μ) > 0 && k >= 0 && k <= 1 &&
μ >= 0 && μ <= 1 && β1 >= 0 && β1 <= 1 &&
γ1 >= 0 && γ1 <= 1 && q >= 0 &&
q <= 1 && γ2 >= 0 && γ2 <= 1;
Flatten[{((γ2 + μ) -
n*κ *(β1*(1 - q) + β2*
q)) - ((k + μ)*(γ1 + μ) - β1*n*
k*(1 - q)*(γ2 + μ)) > 0, con}];
FindInstance[%, {κ, μ, γ1, γ2, β1, β2, k, q}]
(*{{κ -> 0, μ -> 1/2, γ1 -> 0, γ2 -> 0, β1 -> 0, β2 -> 0, k -> 0, q -> 0}}*)
Therefore, the inequality under consideration has a solution.
Though solutions exist, as shown by the other answers, the inequality does not hold under the constraints in general. With N=2000 and all variables zero except $\mu=1$ we can satisfy the constraint, but not the inequality.
Ν = 2000;
vars = {k, μ, γ1, γ2, β1, β2, κ, q};
con = ((k + μ)*(γ1 + μ) - β1*Ν*k*(1 - q)*(γ2 + μ) > 0);
ineq = (((γ2+μ) - Ν*κ (β1*(1-q) + β2*q)) - ((k+μ)*(γ1+μ) - β1*Ν*k*(1-q)*(γ2+μ)) > 0);
Resolve[
ForAll[{k, μ, γ1, γ2, β1, β2, κ, q},
(And @@ (0 <= # <= 1 & /@ vars)),
Implies[con, ineq]], Reals]
sol = First@
FindInstance[
con && (And @@ (0 <= # <= 1 & /@ vars)) && Not[ineq],
{k, μ, γ1, γ2, β1, β2, κ, q}, Reals]
(* {k -> 0, μ -> 1, γ1 -> 0, γ2 -> 0, β1 -> 0, β2 -> 0, κ -> 0, q -> 0} *)
con /. sol (* True *)
ineq /. sol (* False *)
Update 1: After the edits to the question and comments suggesting a different variable range, it's still the case that it does not hold in general and there are solutions which pass the constraints but fail the inequality:
Ν = 2000;
vars = {k, μ, γ1, γ2, β1, β2, q};
con = ((k + μ)*(γ1 + μ) - β1*Ν*k*(1 - q)*(γ2 + μ) > 0);
ineq = ((γ2 + μ) - Ν*k (β1*(1 - q) + β2*q)) -
((k + μ)*(γ1 + μ) - β1*Ν*k*(1 - q)*(γ2 + μ)) > 0;
ranges = (0 < k < 1) && (0 < q < 1) && (And @@ (0 < # & /@ {μ, γ1, γ2, β1, β2}));
sol = First@FindInstance[con && ranges && Not[ineq], vars, Reals]
(* {k -> 1/2048, μ -> 1, γ1 -> 1, γ2 -> 1, β1 -> 1, β2 -> 1, q -> 1/2} *)
con /. sol (* True *)
ineq /. sol (* False *)