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.
 A: It suffices to prove that, for all $x > 1$,
$$\ln x < \frac{x^2 - 1}{2x}.$$
Let $f(x) := \mathrm{RHS} - \mathrm{LHS}$. We have, for all $x > 1$,
$$f'(x) = \frac{(x - 1)^2}{2x^2} > 0.$$
Also, $f(1) = 0$. Thus, $f(x) > 0$ for all $x > 1$.
We are done.
A: 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 $x>1$
A: $\frac{1}{t}$ 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
$$ \ln x < \frac{x-1}{6}\left(1+\frac{1}{x}+\frac{8}{x+1}\right). $$
A: Another way, more geometrical is this one: since, as already pointed out by @MathFail, you have that if $
f(x) = 2x\log x + 1 - x^2$
then $f'(x)=2\log x+2-2x$ you have that your inequality is true if $f'(x)<0$ with $x>1$
This is true if $
\log x < x - 1
$ and this inequality holds since $y=x-1$ is the tangent in $x=1$ to the graphic of $g(x)=\log x$ and this function in concave.
