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The known analytical formula for composite functions is:
f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x
The analytical formula for f [x] cannot be obtained through this method：
In[62]:= Clear["Global`*"]
RSolve[{f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x}, f[x], x] // FullSimplify
Out[63]= RSolve[{1/x + x == f[(1 + x)/Sqrt[x]]}, f[x], x]
and this method is also cannot
Clear["Global`*"]
Simplify /@ (f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x /.
Solve[y == Sqrt[x] + 1/Sqrt[x], x][[1]]) /. {y -> x,
Equal -> Rule}
Clear["Global`*"]
eqn1 = f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x;
sol = Solve[y == Sqrt[x] + 1/Sqrt[x], x] // Quiet
(* {{x -> 1/2 (-2 + y^2 - y Sqrt[-4 + y^2])},
{x -> 1/2 (-2 + y^2 + y Sqrt[-4 + y^2])}} *)
Using first solution,
eqn2 = f[y] == Simplify[x + 1/x /. sol[[1]]] /.
{y -> x, Equal -> Rule}
(* f[x] -> -2 + x^2 *)
Verifying,
eqn1 /. f -> Function[{x}, x^2 - 2] // Simplify
(* True *)
Using the second solution,
eqn3 = f[y] == Simplify[x + 1/x /. sol[[2]]] /.
{y -> x, Equal -> Rule}
(* f[x] -> -2 + x^2 *)
This is identical to the first result.
Assuming f is invertible:
eqn = f[y] == f[Sqrt[x] + 1/Sqrt[x]] == x + 1/x;
Solve[Eliminate[eqn, {x}], f[y]]
InverseFunction::ifun: Inverse functions are being used. Values may be lost for multivalued inverses.
{{f[y] -> -2 + y^2}}
Note: The solution f is invertible on the implicit domain, which is the range of the argument Sqrt[x] + 1/Sqrt[x] (Re[y] >= 0). However, the method works on Solve[Eliminate[f[y] == f[Sqrt[x] + 1/Sqrt[x]] == 2, {x}], f[y]], even though f[y] -> 2 is not invertible.
The warning arises because Solve uses InverseFunction[f][f[u]] to extract the argument u, which is what you see others (in the OP and other answer) doing by hand.
Reduce[Sqrt[x] + 1/Sqrt[x] == x + 1/x]is true when x=1 only. What isfrole here? I am not reading your question right it seems. $\endgroup$fF represents the corresponding relationship of the function $\endgroup$Sqrt[x] + 1/Sqrt[x]and want me to return backx + 1/xthen only whenx=1this is possible. I am not sure what you mean by how to representfin this case. Hopefully someone will have an answer. $\endgroup$