Why Lagrange dual solution gives a different solution

Question

The original problem is

min x

s.t. $x^2 \leq 1$

I modified it to

min x

s.t. $x\leq 1$, $x \geq -1$

And sove its dual problem as:

$max_(\lambda, v)(inf_x(x + \lambda (x-1) + v(-x - 1))) = max_(\lambda, v)(1 - 2v) = 1$

Why the solution is different from the original problem, which the optimum should be -1

Answer 1

$$\begin{align}& \max_{\lambda,v \geq 0} \inf_x x + \lambda(x-1) + v(-x-1) \\ &= \max_{\lambda,v\geq 0} \inf_x (1 + \lambda - v)x - \lambda - v \\ &= \max_{\lambda,v\geq 0} \{ -\lambda-v : 1+\lambda-v=0 \}\end{align}$$ The dual optimal solution is $\lambda=0$, $v=1$, and the optimal value is -1.