$f=0$, yielding a cost of $0$, is a feasible point for your second problem. So if the optimal cost is non-zero then it must be negative.
But then $y=x+\lambda f$ would be an improvement over $x$ in the original problem for all $\lambda>0$. Because of the nature of constraints in the second problem $y$ satisfies the equality and binding non-negativity constraints in the first problem. We can choose $\lambda$ small enough so that it also satisfies the non-binding non-negativity constraints. But then $y$ would be superior to $x$ in the original problem, proving that the latter was not an optimal point.