Positive numbers inequality 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.

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 $f(x)=x^{\log_2 3}$:
$$
\sum_{k=1}^{n}\frac1n f(p_k) \ge f(\frac1n(p_1+\cdots+p_n)).
$$
