Prove $2 x \ln(x) + 1 - x^{2} < 0$, for $x > 1$ I am trying to rigorously show the following bound.
\begin{equation}
2 x \ln(x) + 1 - x^{2} < 0, \text{ for $x > 1$}
\end{equation}
Based on plots, it appears to hold for all $x > 1$.
My concern with showing such bounds is dealing with
the boundary points, in this case $x = 1$, which is not
in the support. I Typically would evaluate at this
point (takes the value 0) and show that the function
is decreasing for all $x > 1$, which should conclude the
proof. To that end, we have that
$f^{\prime}(x) = -2 x + 2 \ln(x) + 2, f^{\prime \prime}(x) = -\frac{2(x - 1)}{x}$.
I'm not sure this is valid here though, since $x = 1$ does is not a valid input for the expression under our constraint $x > 1$. I feel that one needs to be delicate in dealing with $x = 1$. Could anyone please
demonstrate this bound rigorously for my learning?
I'd also appreciate any elementary methods (not necessarily using calculus) to show this to aid with
my understanding.

$\frac{1}{t}$ is a convex function on $\mathbb{R}^+$, hence for any $x>1$
$$ \ln x=\int_{1}^{x}\frac{dt}{t} < \frac{x-1}{2}\left(1+\frac{1}{x}\right)=\frac{x^2-1}{2x}$$
as a consequence of the Hermite-Hadamard inequality. We are just stating something that is geometrically pretty obvious:

By replacing the trapezoid method with Simpson's rule, we can also improve the inequality up to
$$ \ln x < \frac{x-1}{6}\left(1+\frac{1}{x}+\frac{8}{x+1}\right). $$
