Estonian Math Competitions

For $2 \le k \le n$, the GM-HM inequality for the numbers $k, \dots, k, 1$, with $k$ repeated $k-2$ times, gives
$$k - 1 = \frac{(k - 1)^2}{k - 1} = \frac{k(k - 2) + 1}{k - 1} \ge \sqrt[k-1]{k^{k-2}} = \sqrt[k-1]{\frac{k^{k-1}}{k}} = \frac{k}{\sqrt[k-1]{k}}$$
Hence $\sqrt[k-1]{k} \ge \frac{k}{k-1}$ for every $k = 2, 3, \dots, n$, with equality only for $k = 2$. Therefore
$$2 \cdot \sqrt{3} \cdot \sqrt[3]{4} \cdot \dots \cdot \sqrt[n-1]{n} > \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdot \dots \cdot \frac{n}{n-1} = n.$$

Solution 2:
For $2 \le k \le n$, the HM-GM inequality for the numbers $1, \dots, 1, k$ (with $1$ repeated $k-2$ times) gives
$$\sqrt[k-1]{k} \ge \frac{k-1}{k-2 + \frac{1}{k}} = \frac{(k-1)k}{(k-2)k+1} = \frac{(k-1)k}{(k-1)^2} = \frac{k}{k-1},$$
with equality only for $k=2$. We continue as in solution 1.

Solution 3:
From the binomial theorem,
$$\left(\frac{k}{k-1}\right)^{k-1} = \left(1 + \frac{1}{k-1}\right)^{k-1} = \frac{\binom{k-1}{0}}{(k-1)^0} + \frac{\binom{k-1}{1}}{(k-1)^1} + \dots + \frac{\binom{k-1}{k-1}}{(k-1)^{k-1}}.$$
There are $k$ summands, of which $\frac{\binom{k-1}{0}}{(k-1)^0} = \frac{\binom{k-1}{1}}{(k-1)^1} = 1$, and for $1 < i \le k-1$
$$\frac{\binom{k-1}{i}}{(k-1)^i} = \frac{(k-1) \cdots (k-i)}{i! (k-1)^i} < \frac{(k-1) \cdots (k-i)}{(k-1)^i} < 1.$$
Therefore $\left(\frac{k}{k-1}\right)^{k-1} \le \underbrace{1 + 1 + \dots + 1}_{k \text{ times}} = k$, whence $\sqrt[k-1]{k} \ge \frac{k}{k-1}$, with equality only for $k=2$. We continue as in solution 1.

Solution 4:
We show that $k^k > (k+1)^{k-1}$ for $k \ge 2$. The case $k=2$ is obvious. Suppose now that $k^k > (k+1)^{k-1}$ holds for some $k$, and let's prove $(k+1)^{k+1} > (k+2)^k$. Note that $(k+1)^{k+1} \cdot (k+1)^{k-1} = (k+1)^{2k} = (k^2+2k+1)^k > (k^2+2k)^k = (k(k+2))^k = k^k \cdot (k+2)^k$, giving $\frac{(k+1)^{k+1}}{k^k} > \frac{(k+2)^k}{(k+1)^{k-1}}$. This combined with the induction assumption gives
$$(k+1)^{k+1} = \frac{(k+1)^{k+1}}{k^k} \cdot k^k > \frac{(k+2)^k}{(k+1)^{k-1}} \cdot (k+1)^{k-1} = (k+2)^k.$$
We have proven $k^k > (k+1)^{k-1}$, which is equivalent to $\sqrt[k-1]{k} > \sqrt[k]{k+1}$. Therefore $\sqrt[k-1]{k} > \sqrt[k]{n}$ for $k=2, 3, \dots, n-1$, and
$$2 \cdot \sqrt{3} \cdot \sqrt[3]{4} \cdots \sqrt[n-1]{n} > (\sqrt[n-1]{n})^{n-1} = n.$$

Solution 5:
We give another proof for the inequality $k^k > (k+1)^{k-1}$. It is equivalent to $k \cdot \left(\frac{k}{k+1}\right)^{k-1} > 1$, or $k \cdot \underbrace{\frac{k}{k+1} \cdots \frac{k}{k+1}}_{k-1 \text{ times}} > 1$. Note that for $x < k+1$, $x-x \cdot \frac{k}{k+1} = x \cdot (1-\frac{k}{k+1}) = x \cdot \frac{1}{k+1} < 1$. Therefore, multiplication with each factor $\frac{k}{k+1}$ decreases the product by less than 1; cumulatively the product becomes smaller by less than $k-1$. Therefore $k \cdot \left(\frac{k}{k+1}\right)^{k-1} > k-(k-1)=1$.