Prove the function $g(x,y,t)\ge1$

Question

I have the function $$ g(x,y,t)=\frac{(8x^2y^2+f_+(x,y,t)-\cos(2t))(8x^2y^2(1+(x+y)^2)+(x+y)^2(f_-(x,y,t)-\cos(t))+4xy(x+y)\sin(2t))}{64x^4y^4(1+(x+y)^2)} $$ with $$ f_{\pm}(x,y,t) = 1+2x^2+2y^2\pm4xy\cos(2t)-2(x+y)\sin(2t). $$ Is is possible to analytically prove $g(x,y,t)\ge1$? I have numerically checked it for many values of variables (x,y,t). Any suggestion will be helpful.

Just in case, the MMA code of this function is

fp[x_, y_, t_] = 
  1 + 2 x^2 + 2 y^2 + 4 x y Cos[2 t] - 2(x + y)Sin[2t];
fm[x_, y_, t_] = 
  1 + 2 x^2 + 2 y^2 - 4 x y Cos[2 t] - 2(x + y)Sin[2t];
g[x_, y_, 
   t_] = (8 x^2 y^2 + fp[x, y, t] - 
     Cos[2 t]) (8 x^2 y^2 (1 + (x + y)^2) + (x + y)^2 (fm[x, y, t] - 
         Cos[2t]) + 
      4 (x + y) x y Sin[2 t])/(64 x^4 y^4 (1 + (x + y)^2));

Answer 1

This inequality is not true. E.g., $$g\Big(\frac1{10},\frac1{10},\frac\pi4\Big)=-1783.256\ldots\not\ge1.$$

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The OP has changed the problem, thus invalidating the previous answer. After the change, the answer becomes positive. Indeed, the problem is a real algebraic one, since, for any real $u$ and $v$, one can find a real $t$ such that $u=\cos2t$ and $v=\sin2t$ iff $u^2+v^2=1$. Any real algebraic problem can be solved purely algorithmically. In Mathematica, such algorithms are realized via commands Reduce, FindInstance, etc.

In particular, for the function $g$ defined by the Mathematica code in the OP, Mathematica confirms (in about 2.8 sec) that the set of all real triples $(x,y,t)$ such that $g(x,y,t)<1$ is empty: