Inequality from Chapter 5 of the book *How to Think Like a Mathematician* This is from the book How to think like a Mathematician,

How can I prove the inequality
$$\sqrt[\large 7]{7!} &lt; \sqrt[\large 8]{8!}$$

without complicated calculus? I tried and finally obtained just $$\frac 17 \cdot \ln(7!) &lt; \frac 18 \cdot \ln(8!)$$
Your inequality is equivalent to
$$(7!)^8 &lt; (8!)^7$$
divide it by $(7!)^7$, and get
$$7! &lt; 8^7$$
and this is clear, since $$1 \cdots 7 &lt; 8 \cdots 8$$
You have already turned the comparison of two geometric means into the comparison of two arithmetic means.  So consider a more general comparison: show that appending a larger number always raises the geometric mean of a list of positive numbers by showing the effect on the arithmetic mean. Suppose the $x_i$ are real and $x_{n+1}$ is strictly largest.
\begin{equation*}
\begin{split}
(1/(n+1)) \sum_{i=1}^{n+1} x_i &amp;= (1/(n+1)) (x_{n+1} + \sum_{i=1}^{n} x_i)
\\
&amp;=(1/(n+1) (n x_{n+1}/n  + n \sum_{i=1}^{n} x_i / n)
\\
&amp;&gt; (1/(n+1) (\sum_{i=1}^{n} x_i/n  + n \sum_{i=1}^{n} x_i / n)
\\
&amp;= (1/(n+1) ((n+1) \sum_{i=1}^{n} x_i / n)
\\
&amp;= \sum_{i=1}^{n} x_i / n
\end{split}
\end{equation*}

Note that we really only needed $x_{n+1}$ to be larger than the previous mean.