Proving $(n+1)c^{1/(n+1)} - nc^{1/n} \le 1$ from first principles Is it possible to prove that \begin{align*} (n+1)c^{1/(n+1)} - nc^{1/n} \le 1 \qquad c \in \mathbb{R}_+, n \in \mathbb{N} \end{align*} using only elementary techniques? (No calculus, no appeasement to Jensen's inequality). **Motivation** The problem I originally encountered was to prove that, if $x_1, \cdots, x_n > 0$ and $y_n = \sum_{i=1}^{n}{x_i} - n(\prod_{i=1}^{n}{x_i})^{1/n}$, then $(y_n)$ is a monotonically increasing sequence. If we prove this, then our results would imply \begin{align*} A_{n+1} - G_{n+1} \ge \frac{n}{n+1}(A_n - G_n) \end{align*} where $A_n, G_n$ are the arithmetic and geometric means, respectively, of the first $n$ terms. Simple induction would instantly give us a proof of the AM-GM inequality. Other solutions (#33) use the AM-GM inequality itself in proving the statement, while my search for an independent proof not using AM-GM arrived at the inequality you see above.

If $01$, then the inequality (1) will be reversed, but since $1-a$ is negative, the inequality (2) will hold as written.

Rearranging (2), we have $$ (n+1)a^n - na^{n+1} < 1 $$ when $a>0$ and $a\
e 1$; taking $a=c^{1/n(n+1)}$ yields your inequality.