An inequality for exponentials

Question

Let $x_1,\ldots,x_n\in{\bf R}$ satisfying $\sum_{i=1}^nx_i=0$ and $\sum_{i=1}^nx_i^2\le1$. Is it true that for every $r\ge 1$ one has that $\sum_{i=1}^n(r^{x_i}-1)\le r-1$? If so, does anyone have a proof or a reference?

Answer 1

Yes. Note that for $|x| \leq 1$, $$\begin{align} r^x&=1+x\log r+x^2 \sum_{i=2}^{\infty}{\frac{(\log r)^i x^{i-2}}{i!}} \\ & \leq 1+x\log r+x^2 \sum_{i=2}^{\infty}{\frac{(\log r)^i |x|^{i-2}}{i!}} \\ & \leq 1+x\log r+x^2 \sum_{i=2}^{\infty}{\frac{(\log r)^i}{i!}} \\ &=1+x\log r+(r-1-\log r)x^2 \\ \end{align}$$

Note clearly each $|x_i| \leq 1$. Thus $$\sum_{i=1}^{n}(r^{x_i}-1) \leq \log r \sum_{i=1}^{n}{x_i}+(r-1-\log r)\sum_{i=1}^{n}{x_i^2} \leq r-1-\log r \leq r-1$$