Prove that $\log{\frac{\sum_{j=1}^n x_j}{n}}\ge \frac{\sum_{j=1}^n \log x_j}n$ for $j=1,...,n$

Question

Prove that $\log{\frac{\sum_{j=1}^n x_j}{n}}\ge \frac{\sum_{j=1}^n \log x_j}n$ for $j=1,...,n$.

My Work. LHS: \begin{align}\log{\frac{\sum_{j=1}^n x_j}{n}}&=\log{\sum_{j=1}^n x_j}-\log n\\&=\log x_1 +\log{\sum_{j=2}^n\left(1+\frac{x_j}{x_1}\right)}-\log n\\&=\log \left(\frac{x_1}n\right)+\log{\sum_{j=2}^n\left(1+\frac{x_j}{x_1}\right)}\end{align} I don't know if this is the correction to take this problem, but I just tried to take advantage of the properties of a log as best I could

RHS: $$\frac{\sum_{j=1}^n \log x_j}n=\frac 1n(\log x_1+\log x_2+...+\log x_n)=\frac 1n (\log(x_1x_2x_3...x_n))$$ Again, I'm not really sure where I'm going with this, just trying to take advantage of logarithm properties. I feel like I'm missing a logarithm property that can be used to connect the RHS to the LHS, but I'm not sure what...

Answer 1

The function $\log{(x)}$ is strictly concave (calculate the second derivative to see this) in its domain $x>0$. So, for a concave function $φ$, points $x_1,\ldots,x_n$ and positive weights $α_i>0, i=1,\ldots,n$ Jensen's inequality states that $$φ\left(\frac{\sumα_ix_i}{\sum α_i}\right)\ge \frac{\sumα_iφ(x_i)}{\sum α_i}$$ In your case $φ(x_i):=\log{(x_i)}$ and $α_i=1$ for all $i=1,2,\ldots, n$.

Answer 2

Hint: start with AM-GM inequality.