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 $n$.

Answer 1

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.

Answer 2

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.

1. The case of $n = 1$ and $n = 2$ can be proved by brutal force, so we will assume $n \geq 3$.

2. 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.