# Further terms of Taylor expansion for improved inequalities?

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!

Let $$a$$, $$b$$ and $$c$$ be positive numbers such that $$abc=1$$.
Prove that: $$(a+b)^{a+b}+(b+c)^{b+c}+(c+a)^{c+a}\geq4(a+b+c).$$