Let $x\in\mathbb{R}$. Consider the problem
$$\text{minimize}\quad \frac{1}{1+x^2}\quad\text{s.t}\quad x\geq 1$$
Show that the logaritmic primal barrier $B(x,\rho)$ is not bounded from above (for $x\in \Omega$) for any fixed $\rho>0$
My try:
I checked and the limit of $B$ when $x$ goes to infinity is infinity
Now, let
$\quad B(x,\rho)=\frac{1}{1+x^2}+\rho*log(x-1)\quad$ for $\quad\rho>0$,
Let's suppose that $B(x,\rho)$ is bounded from above. Then there exists $b\in \mathbb{R}$ such that $y<b$ for all $y \in B$.
This implies that $$\frac{1}{1+x^2}+\rho*log(x-1)< b$$
I don't see how to solve for $x$ here.
Any suggestions would be great! I'm not sure if contradiction is the right way.