The question at hand is:
Let $(x_i)_{1}^{n}$ be a finite sequence of positive numbers whose mean is
$m = \frac{1}{n}\sum_{i=1}^{n} x_i$
Use the fact that for each positive $t$, we have
$\log t \leq t − 1 $
to show that
$$ \frac{1}{n}\sum_{i=1}^{n} \log(x_i) \leq \log\biggl(\frac{1}{n}\sum_{i=1}^{n} x_i \biggr) = \log m$$
I know that this is a precursor to the AM-GM inequality but I am not allowed to use anything like Jensen's Inequality here, I have repeatedly tried to make the inequality work by applying it to each term in the sequence but have made very little progress. The answer is probably quite straightforward but I would appreciate any guidance here. Thanks
You can apply the inequality $\log t \leq t − 1$ to the values $t = x_i / m$: $$ \begin{align} \left( \frac 1 n\sum_{i=1}^{n} \log(x_i)\right) - \log(m) &= \frac 1 n\sum_{i=1}^{n} \bigl( \log(x_i) - \log (m)\bigr) \\ &= \frac 1 n\sum_{i=1}^{n} \log \left( \frac{x_i}{m}\right) \\ &\le \frac 1 n\sum_{i=1}^{n} \left( \frac{x_i}{m} - 1\right) \\ &= 0 \, . \end{align} $$
Remark: The proof of Jensen's inequality for a concave function $\varphi$ essentially uses that the graph of a concave function lies below its tangent line at a suitable point. For $\varphi(t) = \log (t)$ that would $$ \log(t) \le \varphi(1) + (t-1) \varphi'(1) = t-1 \, . $$ So the above proof of the AM-GM inequality is a proof using Jensen's inequality in disguise.