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.
Best Answer
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.