Prove $2 x \ln(x) + 1 - x^{2} < 0$, for $x > 1$

Question

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.

River Li · Accepted Answer · 2022-07-23T02:36:33.807

3

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.

edited Jul 23 '22 at 02:36

answered Jul 23 '22 at 00:13

River Li

37,323

Thank you, but how do we characterize equality at $x = 1$? This is what I'm hoping to learn to do efficiently such inequalities in my future work. Since the inequality is strict and applies only for $x > 1$ not $x \geq 1$. If we do this, then we can claim $f(x) \geq 0$ for all $x \in [1, \infty)$, with equality iff $x = 1$. – user4687531 Jul 23 '22 at 01:57
I guess you can conclude this, by saying that $f^{\prime}(x) > 0$ (strict inequality), for $x > 1$ above? I just want to be rigorous on iff conditions in inequalities, hence my clarification. – user4687531 Jul 23 '22 at 02:01
@user4687531 I will edit. – River Li Jul 23 '22 at 02:34
@user4687531 I think we can directly use the fact that: If $f'(x) > 0$ on $(a, \infty)$ and $f(a) = 0$, then $f(x) > 0$ on $(a, \infty)$. Or you may prove it use the argument in MathFail's answer. – River Li Jul 23 '22 at 03:06
I think in this case, this argument covers off the boundary case of $x = 1$ quite efficiently. Thanks! – user4687531 Jul 23 '22 at 03:08
@user4687531 You are welcome. – River Li Jul 23 '22 at 03:09

MathFail · Answer 2 · 2022-07-22T22:40:25.247

2

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 $x>1$

edited Jul 22 '22 at 22:40

answered Jul 22 '22 at 22:32

MathFail

21,128

Thank you. However, I believe the same type of issue occurs with establishing $f^{\prime}(x) < 0$, per your claim above. How to deal with this efficiently and rigorously? – user4687531 Jul 22 '22 at 22:48
To prove $f'(x)<0$ strictly, you can do $f''(x)$ and repeat this trick, until exhausts all log terms @user4687531 – MathFail Jul 22 '22 at 22:51
Thank you again. I suspected this was the case. I like your approach, but was hoping to avoid multiple repititions of the same trick. My goal is to learn some efficient 'tricks' to wrap up such inequalities in my future work. – user4687531 Jul 23 '22 at 01:58

Jack D'Aurizio · Answer 3 · 2022-07-23T14:00:31.717

$\frac{1}{t}$ 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

$$ \ln x < \frac{x-1}{6}\left(1+\frac{1}{x}+\frac{8}{x+1}\right). $$

score 0 · Answer 4 · answered Jul 23 '22 at 06:46

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.

Prove $2 x \ln(x) + 1 - x^{2} < 0$, for $x > 1$

4 Answers4