Why is this unbounded below on $\frac{1}{2}$?

Solution 1:

$$g(x_1,x_2,\lambda)=x_1+x_2+x_1x_2+\lambda(x_1^2+x_2^2-1)$$

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 $\lambda\geq \dfrac{1}{2}$.