Innocent-looking family of inequalities

Question

I recognize that one should give "context" to questions, but sometimes it is hard because you could end up writing much and bring in needless flooding of the post. Just to say: the question arises while proving convexity of rational functions from combinatorics. Other than that, it is a fairly complicated background. So, it would nice if someone can pay attention to it.

QUESTION. Let $n\geq1$ be an integer and $0<x<1$ is a real number. Is this inequality true? $$n(2n-1)-2n(2n+1)x+(n+1)(2n+1)x^2+(4n-1)x^{2n-1}\geq0.$$

NOTE. Noticing that the function is clearly convex, i.e. $\frac{d^2}{dx^2}\geq0$, I tried finding a root of the derivative with the intention of showing that the minimum value is positive. That did not work for me. Mind you, the object is not montonic in the variable $n$.

Please still provide context in terms of what you tried, and what the background is. — Calvin Lin, Apr 03 '23 at 18:14
I like your "innocent-looking" qualification. I wonder what could be "guilty-looking" equations. — Jean Marie, Apr 03 '23 at 21:33
@JeanMarie perhaps this inequality with its four sins would be guilty-looking. — Benjamin Wang, Nov 08 '23 at 10:34

River Li · Accepted Answer · 2023-11-08T10:27:38.643

3

Proof. The case of $n = 1, 2, 3$ is verified directly. In the following, assume that $n \ge 4$.

If $0 < x \le \frac{n}{n + 2}$, we have \begin{align*} &n(2n-1)-2n(2n+1)x+(n+1)(2n+1)x^2 \\ ={}& (n+1)(2n+1)\left(\frac{n}{n+1} - x\right)^2 - \frac{n}{n+1}\\ \ge{}& (n+1)(2n+1)\left(\frac{n}{n+1} - \frac{n}{n + 2}\right)^2 - \frac{n}{n+1}\\ ={}& \frac{n(n - 4)}{(n+2)^2}\\ \ge{}& 0. \end{align*}
If $\frac{n}{n + 2} < x < 1$, we have \begin{align*} &n(2n-1)-2n(2n+1)x+(n+1)(2n+1)x^2+(4n-1)x^{2n-1}\\ ={}& n(2n-1)-2n(2n+1)x+(n+1)(2n+1)x^2+ (4n-1)\cdot \frac{n}{4n^2 + 3n - 1}\\[6pt] &\qquad + (4n-1)\left(x^{2n-1} - \frac{n}{4n^2 + 3n - 1}\right)\\[6pt] ={}& \frac{(2n+1)(nx + x - n)^2}{n+1} + (4n-1)\left(x^{2n-1} - \frac{n}{4n^2 + 3n - 1}\right)\\ \ge{}& \frac{(2n+1)(nx + x - n)^2}{n+1} + (4n-1)\left(\left(\frac{n}{n + 2}\right)^{2n-1} - \frac{n}{4n^2 + 3n - 1}\right). \end{align*}

It suffices to prove that $$\left(\frac{n}{n + 2}\right)^{2n-1} - \frac{n}{4n^2 + 3n - 1} \ge 0$$ or $$\frac{4n^2 + 3n - 1}{n} \ge (1 + 2/n)^{2n - 1}$$ or $$\frac{4n^2 + 3n - 1}{n}\cdot (1 + 2/n) \ge (1 + 2/n)^{2n}.$$

It is easy to prove that $$\frac{4n^2 + 3n - 1}{n}\cdot (1 + 2/n) \ge \mathrm{e}^4\left(1 - \frac13\cdot \frac{2}{n}\right)^4.$$

Thus, it suffices to prove that $$\mathrm{e}\left(1 - \frac13 \cdot \frac{2}{n}\right) \ge (1 + 2/n)^{n/2}.$$ Letting $x = 2/n \in (0, 1/2]$, it suffices to prove that $$\mathrm{e}(1 - x/3) \ge (1 + x)^{1/x}.$$ Let $f(x) := \mathrm{e}(1 - x/3) - (1 + x)^{1/x}$. It is easy to prove that $x\mapsto (1 + x)^{1/x}$ is convex on $(0, 1/2]$. Thus, $f(x)$ is concave. Also, we have $f(1/2) > 0$ and $\lim_{x\to 0^{+}} f(x) = 0$. Thus, we have $f(x) \ge 0$ on $(0, 1/2]$.

We are done.

edited Nov 08 '23 at 10:27

answered Apr 04 '23 at 12:27

River Li

37,323

This is nice. The last inequality needs proof though. – T. Amdeberhan Apr 04 '23 at 13:47
1

@T.Amdeberhan It is smooth e.g. by AM-GM for $Ax^2 + C - Bx \ge 2\sqrt{AC} x - Bx\ge 0$. – River Li Apr 04 '23 at 14:03
Good point. Just to be picky: $(1+2/n)^{n/2}\leq e(1-1/n)$ is not true; $(1+2/n)^{n/2}\leq e$ is true, for any $n\geq1$, in which case the last inequality you had (via AGM) is valid for $n\geq12$. Of course, one has to verify for $n=4,\dots,11$ (a few cases). – T. Amdeberhan Apr 04 '23 at 15:25
@T.Amdeberhan Nice point. I will edit it. – River Li Apr 05 '23 at 11:04
@T.Amdeberhan I edited it. Is it OK now? – River Li Apr 05 '23 at 11:14
thank you. It is correct, but the AGM inequality comes up with nasty expressions. Also, if you don't mind, can you explain the 1st inequality (the case $x<n/(n+2)$)? It is easier but I wish to see your approach. – T. Amdeberhan Apr 05 '23 at 14:51
Yes, it is not easy by hand. 2) Since $n(2n-1)-2n(2n+1)x+(n+1)(2n+1)x^2 \ge 0$ is equivalent to $x < x_1$ or $x > x_2$, it suffices to prove that $\frac{1}{1 + 2/n} < x_1$. I think we may have simpler proof.

River Li

Apr 05 '23 at 15:02

score 2 · Answer 2 · answered Nov 07 '23 at 19:09

Let $f(x)$ denote the left-hand side of the inequality. Substituting $x = \frac{n+t}{n+1}$ for $-n < t < 1$,

\begin{align*} f\left(\frac{n+t}{n+1}\right) &= \frac{(2n + 1)t^2 - n}{n+1} + (4n - 1)\left(\frac{n+t}{n+1}\right)^{2n-1}. \end{align*}

Before moving forward, let us make some assumptions.

The case of $n = 1$ and $n = 2$ can be proved by brutal force, so we will assume $n \geq 3$.
The above substitution shows that $f(\frac{n+t}{n+1})$ is always non-negative for $|t| \geq \frac{1}{\sqrt{2}}$. Hence, it suffices to consider the case $|t| < \frac{1}{\sqrt{2}}$ and we assume so hereafter.

By using the inequality $ x \geq e^{(x-1)/x} $, $0 < x < 1$, we get

\begin{align*} \left(\frac{n+t}{n+1}\right)^{2n-1} \geq e^{-(2n-1)\frac{1-t}{n+t}} \geq e^{-2(1-t)}. \end{align*}

From this,

\begin{align*} f\left(\frac{n+t}{n+1}\right) \geq 0 &\impliedby \frac{(2n + 1)t^2 - n}{n+1} + (4n - 1)e^{-2(1-t)} \geq 0 \\ &\iff (4n - 1)\frac{n+1}{2n+1} \geq \left(\frac{n}{2n+1} - t^2 \right) e^{2(1-t)} \\ &\impliedby \inf_{n\geq 3} (4n - 1)\frac{n+1}{2n+1} \geq \sup_{n\geq 3}\left(\frac{n}{2n+1} - t^2 \right) e^{2(1-t)} \\ &\iff \frac{44}{7} \geq \left(\frac{1}{2} - t^2 \right) e^{2(1-t)}. \end{align*}

Now the maximum of RHS for $|t| < \frac{1}{\sqrt{2}}$ can be easily located. Indeed, it turns out that RHS is maximized at $t = \frac{1}{2} (1-\sqrt{3})$ with maximum value of $\frac{1}{2} (\sqrt{3}-1) e^{\sqrt{3}+1} \approx 5.62375$, which is smaller tha $\frac{44}{7} \approx 6.28571$. Therefore the desired claim follows.

Innocent-looking family of inequalities

2 Answers2