It is well known that for a finite set of real numbers $(a_1, \dots, a_n)$ the mean value is upper bounded by the maximum: $\bar{a} = \frac{1}{n}\sum_{k=1}^{n}a_k \le \max(a_1,\dots,a_n) = a_{\rm max}$. A straightforward proof would be
$$ \frac{1}{n}\sum_{k=1}^na_k \le \frac{1}{n}\sum_{k=1}^{n}a_{\rm max} = a_{\max}. $$
Given that this question does not appear to have been asked on the site before, and to illustrate the power of simple inductive arguments, I am interested in the collection of proofs by mathematical induction that members on the site come up with.
Define $\bar a_0=0$ and $\bar a_k$ for $1\leqslant k\leqslant n$ by $$ \bar a_k = \frac{k-1}k \bar a_{k-1} + \frac1k a_k. $$ Then $$ \bar a_1 = 0\cdot 0 + 1\cdot a_1 = \frac11\sum_{j=1}^1 a_j, $$ and assuming that $\bar a_k = \frac1k\sum_{j=1}^k a_j$, then \begin{align} \bar a_{k+1} &= \frac k{k+1}\bar a_k + \frac 1{k+1} a_{k+1}\\ &= \frac k{k+1}\cdot\frac1k\sum_{j=1}^k a_j + \frac 1{k+1} a_{k+1}\\ &= \frac1{k+1}\sum_{j=1}^{k+1}a_j, \end{align} so $\bar a_k$ is a valid inductive definition of the mean. Similarly, define $a^*_0=a_1$ and $$a^*_k=\max\{a^*_{k-1}, a_k\}$$ for $1\leqslant k\leqslant n$. Then $a^*_1 = \max\{a_1,a_1\}=a_1$ and assuming that $a^*_k = \max\{a_j : 1\leqslant j\leqslant k\}$, we have \begin{align} a^*_{k+1} &= \max\{a^*_k, a_{k+1}\}\\ &= \max\{\max\{a_j : 1\leqslant j\leqslant k\}, a_{k+1} \}\\ &= \max\{a_j : 1\leqslant j\leqslant k+1\}, \end{align} so $a^*_k$ is a valid inductive definition of the max. Now, $a^*_k\geqslant a^*_{k-1}$ and $a^*_k\geqslant a_k$, and $\bar a_1 = a_1\leqslant a_1^*$. Assuming $\bar a_k\leqslant a^*_k$ for some $1\leqslant k\leqslant n-1$, we have \begin{align} \bar a_{k+1} &= \frac k{k+1} \bar a_k +\frac1{k+1} a_{k+1}\\ &= \frac 1{k+1}(k\bar a_k + a_{k+1})\\ & \leqslant \frac 1{k+1}(k a^*_k + a^*_{k+1})\\ & \leqslant \frac 1{k+1}(k a^*_{k+1} + a^*_{k+1})\\ &= a^*_{k+1}, \end{align} thus proving that the mean is less than the maximum.