This is a follow-up question to Recognize that the function is unbounded below where we got the following optimization problem to solve with the concept of duality.
$$\min_{x_1, x_2} x_1+x_2+x_1x_2$$ s.t. $$x_1^2+x_2^2 = 1$$
So it shows that the duality problem is:
$$\max_{\lambda} D(\lambda)= \max_{\lambda}\Big\{-\frac{1}{1+2\lambda}-\lambda\Big\},$$
The textbook now states that the duality function $-\frac{1}{1+2\lambda}-\lambda$ is unbounded below when $\lambda < \frac{1}{2}$. And indeed, if you allow a value of $\lambda$ to go to $-\infty$, then the dual function can be maximized to $\infty$. So you have to make a "cut" at some point, in this case $\frac{1}{2}$.
But my question is: why $\frac{1}{2}$ ?? Why is the duality function unbounded below when $\lambda < \frac{1}{2}$ and not when e.g. $<\frac{1}{4}$ ? I can make the same case with $\lambda < \frac{1}{4}$ and use $\frac{1}{4}$ as $\lambda$, which would make my duality function larger than when I use $\frac{1}{2}$.
But the textbook clearly states that the duality function is unbounded below when less than $\frac{1}{2}$...
$$g(x_1,x_2,\lambda)=x_1+x_2+x_1x_2+\lambda(x_1^2+x_2^2-1)$$
Let's check if the minimum $g(x_1,x_2,\lambda)$ exists at contant $\lambda$. This can be done by representing $g$ in some convenient form.
If $\lambda \neq 0, 4\lambda^2\neq 1$:
$$g=\lambda x_1^2+x_1(x_2+1)+\lambda x_2^2+x_2-\lambda=\lambda \left(x_1^2+2x_1 \dfrac{x_2+1}{2\lambda}+\dfrac{x_2^2+2x_2+1}{4\lambda^2}\right)+\\+ \dfrac{4\lambda^2x_2^2+4\lambda x_2-4\lambda^2-x_2^2-2x_2-1}{4\lambda}=\lambda\left(x_1+\dfrac{x_2+1}{2\lambda}\right)^2+\dfrac{4\lambda^2-1}{4\lambda}x_2^2+\dfrac{2\lambda-1}{2\lambda}x_2-\dfrac{4\lambda^2+1}{4\lambda}=\\=A(x_1+Bx_2+C)^2+E(x_2+F)^2+G$$
$A=\lambda$, $E=\dfrac{4\lambda^2-1}{4\lambda}$. Expression is bounded below when $A\geq 0$, $E\geq 0$. This leads to $\lambda> \frac{1}{2}$.
Expression is not bounded below at $\lambda<\frac{1}{2}$ such that $\lambda \neq 0, 4\lambda^2\neq 1$. Cases $\lambda=0$, $\lambda=\pm\dfrac{1}{2}$ should be considered separately.
At $\lambda=-\dfrac{1}{2}$: $$g(x_1,x_2,-1/2)=x_1+x_2-\frac{1}{2}(x_1-x_2)^2+\frac{1}{2}$$ is not bounded, because taking $x_1=x_2=a/2$ one can make $g(a/2,a/2,-1/2)=a+\frac{1}{2}$ equal to any real number.
At $\lambda=0$: $$g(x_1,x_2,0)=x_1+x_2+x_1x_2$$ is not bounded because taking $x_1=a$, $x_2=0$ one can make $g(a,0,0)=a$ equal to any real number $a$.
At $\lambda=\dfrac{1}{2}$: $$g(x_1,x_2,1/2)=x_1+x_2+\frac{1}{2}(x_1+x_2)^2-\frac{1}{2}= \frac{1}{2}(x_1+x_2+1)^2-\frac{1}{2}\geq -1$$
is bounded below. Minimum value of $g$ is $-1$.
At $\lambda>\frac{1}{2}$: point of minimal $g(x_1,x_2,\lambda)$ could be found from system $Ax_1+Bx_2+C=0$, $Ex_2+F=0$. Also it could be found by taking derivatives of $g(x_1,x_2)$. The result is $g(x_1,x_2,\lambda)\geq -\dfrac{1}{2\lambda+1}-\lambda$. One can combine this result with case $\lambda=\dfrac{1}{2}$ and write it as $D(\lambda)=-\dfrac{1}{2\lambda+1}-\lambda$ at $\lambda\geq \dfrac{1}{2}$.