Further terms of Taylor expansion for improved inequalities?

Question

Background: We know the classical exponential inequality \begin{align*} 1 + x \le e^x \qquad \text{or} \qquad x \le e^{x - 1} \end{align*} Taking $x_i = y_i/\overline{y}$, where $y_i \ge 0$ and $\overline{y}$ is the arithmetic mean, and multiplying out the second form, we yield \begin{align*} \frac{\prod_{i=1}^{n}y_i}{\overline{y}^n} \le \exp\left(\frac{\sum_{i=1}^{n}y_i}{\overline{y}} - n\right) = 1 \end{align*} And this yields a proof of the arithmetic-means and geometric-means (AM-GM) inequality.

An improved exponential inequality: We also have the exponential inequality \begin{align*} 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \le e^x \end{align*} If we substitute $y = 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$, we get the equivalent form \begin{align*} y \le e^{g(y)} \end{align*} where \begin{align*} g(y) = \left(\sqrt{9y^2 - 6y + 2} + 3y - 1\right)^\frac{1}{3} - \left(\sqrt{9y^2 - 6y + 2} + 3y - 1\right)^{-\frac{1}{3}} - 1 \end{align*} is the inverse function expressing $x$ in terms of $y$.

Can we derive other inequalities? From here, there are many directions we may take with our given inequality, but due to the non-linear nature of $g(y)$, a direct trick of normalization as in the AM-GM proof will yield no immediate discoveries.

This post is meant to be very open to discuss possible inequality sharpenings from the cubic-Taylor expansion inequality above. Some observations:

1. $(g(y) + 1)^3 = -3(g(y) + 1)$
2. $[6y-2]_+^\frac{1}{3} - [6y-2]_+^{-\frac{1}{3}} - 1 < g(y) \le ([6y-2]_+ + 1)^\frac{1}{3} - ([6y-2]_+ + 1)^{-\frac{1}{3}} - 1$, where $[x]_+ = \max(x, 0)$.

Have fun!

Answer 1

By your idea we can get for example the following inequality.

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$.

Prove that: $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}\geq4(a+b+c).$$