How to show that the following two optimization problems are equivalent?

The first problem is $$\max. ~~g(\lambda)-x$$ $$s.t. ~~ x\geq 0, \lambda \geq 0, a-\lambda b +x \geq 0$$ where $$g(\lambda)$$ is an increasing function of $$\lambda$$ and $$a,b$$ are some positive constants and the optimization variables are $$x,\lambda$$.

I want to know how the above problem is equivalent to the following problem $$\min. ~~g(\lambda)+x$$ $$s.t. ~~ x\geq 0, \lambda \geq 0, x \geq a-\lambda b.$$ At this moment, I only understand the change in sign before $$x$$ in the objective function of second optimization problem. But I do not understand the rest of the changes. Any help in this regard will be much appreciated.

From the constraints in the first problem, we have $$$$-x \leq 0, \quad -x \leq a-\lambda b.$$$$ Thus $$$$g(\lambda) - x \leq g(\lambda) + \min(0, a-\lambda b).$$$$ Then \begin{aligned} \sup_{\lambda,x} \left\{g(\lambda) -x \right\} &= \max(\sup_{a-\lambda b \ge 0} \left\{g(\lambda) \right\}, \sup_{a-\lambda b \leq 0} \left\{g(\lambda) + a-\lambda b\right\}) \\ &= \max(g(\frac{a}{b}), \sup_{a-\lambda b \leq 0} \left\{g(\lambda) + a-\lambda b\right\}), \end{aligned} The second equality is derived by the fact that $$g$$ is increasing.
Similarly, for the second problem, we have \begin{aligned} \inf_{\lambda, x} \left\{g(\lambda)+x\right\} &= \min(\inf_{a-\lambda b \leq 0} \{g(\lambda)\}, \inf_{a-\lambda b \ge 0} \{g(\lambda)+a-\lambda b\}) \\ &= \min(g(\frac{a}{b}), \inf_{a-\lambda b \ge 0, \lambda \ge 0} \{g(\lambda)+a-\lambda b\}), \end{aligned}
This two problems are not equivalent. And we need some more conditions, for example, suppose $$g(\lambda) - \lambda b$$ is decreasing in terms of $$\lambda$$. In this sense, we can easily know that $$$$\sup_{\lambda,x} \left\{g(\lambda) -x \right\} = g(\frac{a}{b}), \quad \inf_{\lambda, x} \left\{g(\lambda)+x\right\} = g(\frac{a}{b}).$$$$
On the other hand, if $$g(\lambda) - \lambda b$$ is NOT decreasing in terms of $$\lambda$$, we can find some counterexamples:
• $$a=b=1$$ and $$g(\lambda) = \lambda^3$$. The maximum of the first problem is $$+\infty$$ and the minimum of the second problem is $$1$$.
• $$a=b=1$$ and $$g(\lambda) = 2\lambda$$. The maximum of the first problem is $$+\infty$$ and the minimum of the second problem is $$1$$.