I have this inequality
$$-x^2-\left(x^4 \cos (3 x)+9 x^2\right)\;\leq \sqrt{f (x)}\;\leq\; \left(x^4 \cos (3 x)+9 x^2\right)-x^2$$ where $\;f (x)=x^4 \cos (\pi x)+x-\cos (x),\;$ and $x>0$. As can be seen, the functions on the LHS and RHS of the inequality always give real results, but, the function $f(x)$ for some values of $x$ can be negative and hence, for those values, $\sqrt{f (x)}$ will have imaginary parts as well.
My question is since the LHS and RHS are always real, can I claim that $f(x)$ must be necessarily positive?
In other words, is this inequality valid only for $f(x)>0$?
In answer to the OP's official question, not only can you claim that $f(x)$ must be non-negative (note the inclusion of $0$), you must claim it. The only way not to do so would be to redefine the inequality symbol in some weird way.
In other words, for the OP's function, let
$$I_f=\left\{x: -x^2-\left(x^4 \cos (3 x)+9 x^2\right)\;\leq \sqrt{|f (x)|}\;\leq\; \left(x^4 \cos (3 x)+9 x^2\right)-x^2\right\}$$
and
$$N_f=\left\{x:f(x)\ge0\right\}$$
Then the solution to the OP's inequality is $I_f\cap N_f$. Neither $I_f$ nor $N_f$ appears at all easy to solve for explicitly; it would be remarkable if their intersection were any easier.