Positive numbers inequality

Question

Let $ x_1, x_2, \ldots, x_n \; \in \; \mathbb{R}_{+} $ so that $ x_1 \geq x_2 \geq \cdots \geq x_n $

We also know that: $$ \frac{1}{2^{x_1}} + \frac{1}{2^{x_2}} + \cdots + \frac{1}{2^{x_n}} = 1 $$

Prove the following inequality: $$ \frac{1}{3^{x_1}} + \frac{2}{3^{x_2}} + \cdots + \frac{n}{3^{x_n}} \geq \frac{n + 1}{2n} $$

I've tried to prove it several times but I haven't succeeded. I would really appreciate if you can help me.

Answer 1

Following Michael Hardy's comment, the problem is equivalent to showing $$\sum_{k=1}^n k\cdot p_k^{\log_2 3}\ge \frac{n+1}{2n}$$ if $0<p_1\le \cdots\le p_n$ and $p_1+\cdots+p_n=1$. This can be proven as follows:

\begin{align*} &\sum_{k=1}^n k\cdot {p_k}^{\log_2 3}\\ &\ge \sum_{k=1}^n \frac{n+1}{2}\cdot {p_k}^{\log_2 3}\\ &= \frac{n+1}{2} \sum_{k=1}^n {p_k}^{\log_2 3}\\ &\ge \frac{n+1}{2} n\left(\frac1n\right)^{\log_2 3}\\ &\ge \frac{n+1}{2} n\left(\frac1n\right)^2\\ &=\frac{n+1}{2n}. \end{align*}

1st line to 2nd line: For $1\le k\le \lfloor\frac{n+1}{2}\rfloor$, $$ k {p_k}^{\log_2 3}+(n+1-k) {p_{n+1-k}}^{\log_2 3} \ge \frac{n+1}{2} {p_k}^{\log_2 3}+\frac{n+1}{2} {p_{n+1-k}}^{\log_2 3}. $$

3rd line to 4th line: apply Jensen's inequality to a convex function $f(x)=x^{\log_2 3}$: $$ \sum_{k=1}^{n}\frac1n f(p_k) \ge f(\frac1n(p_1+\cdots+p_n)). $$