Suppose I have the following constrained optimization problem
$$max \quad f(x) \quad s.t. \quad g(x)=a$$
whose solution is denoted by $x^{*}$. I want to prove that this is the solution to the above mentioned proble in a diffferent way.
So let's consider the new function $z(g(x))$, which is logically equivalent to the initial $f(x)$ and has the same $x$ that maximizes it, but it has just be re-arranged to express the problem as a function of $g(x)$ instead of $x$ so that we could incorporate the effect of the constraint.
Assume that we take the derivative of $z(g(x))$ with respect to $g(x)$ and set it equal to $0$ ($\frac{\delta z}{\delta g} =0$ ) to find the optimal $g(x)$. Under the optimum I have the condition $g(x)-b=0$. If I now substitute the solution of the initial constrained problem $x^{*}$ into the expression above, I find that $g(x^{*})=a=b$. Is that enough to prove that $x^{*}$ is the optiumum for the intial constrained problem?
In other words, can I re-express an optimization problem like $$max \quad f(x) \quad wrt \quad x \quad st \quad g(x)=a$$ as a new optimization problem $$max \quad z(g(x)) \quad wrt \quad g(x) \quad st \quad g(x)=a$$ so that i can put $\frac{\delta z}{\delta g} =0$ find the expression for the optimal $g(x)$ like $g(x)-b=0$ and later solve $g(x)-b$ for $x$ to find the optimal $x^{*}$ (provided that $x^{*}$ is such that $g(x^{*})=a$)?
Basic example: consider the minimization of the function $y=f(x)=(x-2)^{2}$ st $(x-2)^{0.5}>=0$, which is minimized for x=2. Then put g(x)=(x-2)^{0.5} and rewrite z(g(x))=(g(x)^{2})^{2}=g(x)^{4}: the g(x) that minimizes z(g(x)) is 0. Now remember that g(x)=(x-2)^{0.5}, therefore solving g(x)=0 for x gives x=2, which indeed is the optimum for the original problem
Clearly proving that it is NOT right to do so is equivalent to proving that there exist cases where solving g(x)=a=b (or, for inequality constraints, g(x)=b>=a) for x gives results that are different from the optimal x for the initial constrained problem