How can we prove the following product-sum inequality involving e?

Question

Take any sequence $x_1, \dots, x_k$ of positive reals such that $\sum_{i=1}^{k-1} x_i < 1$ and $\sum_{i=1}^k x_i \ge 1$. I suspect that the following inequality holds: $$ \prod_{i=1}^k (1 + x_i) \le e \sum_{i=1}^k x_k. $$

This didn't seem very hard at first, but I can't quite get the desired inequality. Here's my best attempt so far. By AM-GM inequality, we have $$ \prod_{i=1}^k (1 + x_i) = (1 + x_k) \prod_{i=1}^{k-1} (1 + x_i) \le (1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1}. $$ By a standard exponential function inequality, we then have $$ (1 + x_k) \left( 1 + \frac{1}{k-1} \sum_{i=1}^{k-1} x_i \right)^{k-1} < (1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right). $$ Finally, by the first constraint on the sequence, we get $$ (1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right) \le e (1 + x_k). $$

This is very close to what I want, but not quite there. I'm assuming there's a trick somewhere that I'm missing. Any help would be appreciated.

Answer 1

Starting with the second step above, we want to show that $$ (1 + x_k) \exp \left( \sum_{i=1}^{k-1} x_i \right) \le e \sum_{i=1}^k x_i, $$ which is equivalent to $$ (1 + a) \exp(b) \le e (a + b), $$ for $0 \le b < 1 \le a + b$.

I tried using polynomial upper bounds on $\exp(b)$ for $0 \le b < 1$, but none of the bounds I tried were tight enough. I suspect that no finite polynomial upper bound on $\exp(b)$ is sufficient to get the desired inequality.

Ultimately, I overlooked a simple solution. First, rewriting it as $f(a, b) = e(a + b) - (1 + a) \exp(b)$, we want to show that $f(a, b) \ge 0$ for all $0 \le b < 1 \le a + b$. This function is smooth and has a saddle point at $b = 1$, $a = 0$ for which $f(a, b) = 0$. Then it's enough to show that $f$ is weakly increasing in $a$ and in $a - b$ for all $a, b$ in the domain. We can check that this is true by differentiating.

Answer 2

If $a\ne b$, $$ \begin{align} \left(1+\frac{a+b}2\right)^2 &=(1+a)(1+b)+\left(\frac{a-b}2\right)^2\tag{1a}\\ &\gt(1+a)(1+b)\tag{1b} \end{align} $$ Induction and $(1)$ shows that for a given sum of $x_k$, the product of $1+x_k$ is maximized when all of the $x_k$ are equal (if two $x_k$ are different, replace them with their average to get greater product while maintaining the sum). That is, $$ \prod_{k=1}^n(1+x_n)\le\left(1+\frac1n\sum_{k=1}^nx_n\right)^n\tag2 $$

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Suppose that $$ \sum_{k=1}^{n-1}x_k=\alpha\lt1\tag3 $$ and $$ \alpha+x_n\ge1\tag4 $$ then $(2)$ and $(3)$ imply that the greatest product of $1+x_k$ is when all the $x_k$ are equal; that is, $$ \begin{align} \prod_{k=1}^n(1+x_k) &=\left(\prod_{k=1}^{n-1}(1+x_k)\right)(1+x_n)\tag{5a}\\ &\le\left(1+\frac\alpha{n-1}\right)^{n-1}(1+x_n)\tag{5b} \end{align} $$ Consider $$ f(\alpha)=e^\alpha(1+x_n)-e(\alpha+x_n)\tag6 $$ Since $f''(\alpha)=e^\alpha(1+x_n)\ge0$, $f$ is convex. Furthermore, $(3)$ and $(4)$ constrain $1-x_n\le\alpha\lt1$.

Since $x_n\gt0$, $$ \begin{align} f(1-x_n) &=e^{1-x_n}\overbrace{(1+x_n)}^{\large\lt e^{x_n}}-e(1-x_n+x_n)\tag{7a}\\ &\lt0\tag{7b} \end{align} $$ and $$ \begin{align} f(1) &=e(1+x_n)-e(1+x_n)\tag{8a}\\ &=0\tag{8b} \end{align} $$ for $1-x_n\le\alpha\lt1$, we have $$ \begin{align} \prod_{k=1}^n(1+x_k)-e\sum_{k=1}^nx_k &\le\left(1+\frac\alpha{n-1}\right)^{n-1}(1+x_n)-e(\alpha+x_n)\tag{9a}\\ &\le e^\alpha(1+x_n)-e(\alpha+x_n)\tag{9b}\\[9pt] &=f(\alpha)\tag{9c}\\[9pt] &\lt0\tag{9d} \end{align} $$ Therefore, $$ \prod_{k=1}^n(1+x_k)\lt e\sum_{k=1}^nx_k\tag{10} $$