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.

Let $f(x)=2x\log x+1−x^2$, so $f'(x)=2\ln(x)+2-2x<0$ for $x\in (1,\infty)$
Assume there is another point $x_1$ on $(1,\infty)$, such that
$f(x_1)=0$, since $f(1)=0$, by Mean Value theorem, there exists $x_2\in (1,x_1)$, such that $f'(x_2)=0$, which contradicts with the fact $f'(x)<0$ for all $x\in (1,\infty)$
This means $f(x)$ will stay always below x-axis and never cross x-axis (by continuity). Therefore, $f(x)<0$ strictly for $x>1$
