Let $a_{1}, a_{2}, \ldots$ be a sequence of positive numbers such that for any positive $n$,

$$a_{1}+a_{2}+\ldots+a_{n}=\frac{1}{a_{n}}$$

Prove that

$$\frac{1}{\sqrt{2 n-1}} \leq a_{n} \leq \frac{1}{\sqrt{n}}$$

I. Solution. Let $S_{n}=a_{1}+a_{2}+\ldots+a_{n}=1 / a_{n}$. We will show that

$$\sqrt{n} \leq S_{n} \leq \sqrt{2 n-1} \quad (\text{this is equivalent to the statement). }$$

We will prove this by induction. For $n=1$, from the first equation, $a_{1}=1 / a_{1}$, so $a_{1}=1, S_{1}=1$, and thus $\sqrt{1} \leq S_{1} \leq \sqrt{2 \cdot 1-1}$ is indeed true.

Let $k>1$ and assume that (3) holds for $k$:

$$\sqrt{k} \leq S_{k} \leq \sqrt{2 k-1}$$

We will show that it also holds for $(k+1)$. For this, we first express $S_{k+1}$ in terms of $S_{k}$. Write the $(k+1)$-th equation:

$$\begin{gathered} a_{1}+a_{2}+\ldots+a_{k}+a_{k+1}=S_{k}+a_{k+1}=\frac{1}{a_{k+1}}, \text { hence } \\ a_{k+1}^{2}+S_{k} a_{k+1}-1=0 \end{gathered}$$

The roots of the equation $x^{2}+S_{k} x-1=0$ are:

$$\frac{-S_{k}+\sqrt{S_{k}^{2}+4}}{2} \text { and } \frac{-S_{k}-\sqrt{S_{k}^{2}+4}}{2}$$

The second root is negative, while $a_{k+1}$ is positive, so

$$a_{k+1}=\frac{-S_{k}+\sqrt{S_{k}^{2}+4}}{2} \quad \text { and thus } \quad S_{k+1}=S_{k}+a_{k+1}=\frac{S_{k}+\sqrt{S_{k}^{2}+4}}{2} .$$

From the induction hypothesis (4), we have

$$\frac{\sqrt{k}+\sqrt{k+4}}{2} \leq S_{k+1} \leq \frac{\sqrt{2 k-1}+\sqrt{2 k+3}}{2}$$

We will show that $(\sqrt{k}+\sqrt{k+4}) / 2 \geq \sqrt{k+1}$, which implies the left side of the statement: $\sqrt{k+1} \leq S_{k}$. Indeed, multiplying by 2 and squaring both sides, we get:

$2 k+4+2 \sqrt{k(k+4)} \geq 4 k+4$, which is true because $\sqrt{k(k+4)} \geq k$.

Similarly, the other inequality to be proven follows if we show that

$$\frac{\sqrt{2 k-1}+\sqrt{2 k+3}}{2} \leq \sqrt{2 k+1}$$

Again, multiplying by 2 and squaring both sides, we get:

$$4 k+2+2 \sqrt{4 k^{2}+4 k-3} \leq 8 k+4$$

which is true because

$$\sqrt{4 k^{2}+4 k-3} \leq \sqrt{4 k^{2}+4 k+1}=2 k+1$$

Thus, we have shown that

$$\sqrt{k+1} \leq S_{k+1} \leq \sqrt{2 k+1}=\sqrt{2(k+1)-1}$$

and we have proven the statement.

II. Solution. Let $b_{n}=\left(\frac{1}{a_{n}}\right)^{2}$. We will show that

$$n \leq b_{n} \leq 2 n-1$$

Write equation (1) for two consecutive indices:

$$a_{1}+a_{2}+\ldots+a_{k}+a_{k+1}=\frac{1}{a_{k+1}}, \quad a_{1}+a_{2}+\ldots+a_{k}=\frac{1}{a_{k}}$$

and subtract the second from the first:

$$a_{k+1}=\frac{1}{a_{k+1}}-\frac{1}{a_{k}}$$

From this, express $1 / a_{k}$:

$$\frac{1}{a_{k}}=\frac{1}{a_{k+1}}-a_{k+1}$$

and square both sides:

$$\begin{gathered} \left(\frac{1}{a_{k}}\right)^{2}=\left(\frac{1}{a_{k+1}}\right)^{2}+a_{k+1}^{2}-2, \quad \text { or } \\ b_{k}=b_{k+1}+a_{k+1}^{2}-2 \end{gathered}$$

Since

$$\frac{1}{a_{k+1}}=a_{1}+\ldots+a_{k+1}>a_{k+1}, \quad \text { it follows that } \quad 1>a_{k+1}^{2}$$

From this, we get

$$\begin{gathered} b_{k+1}-b_{k}=2-a_{k+1}^{2} > 1 \quad \text { follows. } \end{gathered}$$

Finally, since $b_{1}=1$,

$$\begin{aligned} & b_{n}=\left(b_{n}-b_{n-1}\right)+\left(b_{n-1}-b_{n-2}\right)+\ldots+\left(b_{2}-b_{1}\right)+b_{1} \geq (n-1) \cdot 1+1=n \quad \text { and } \\ & b_{n}=\left(b_{n}-b_{n-1}\right)+\left(b_{n-1}-b_{n-2}\right)+\ldots+\left(b_{2}-b_{1}\right)+b_{1} \leq (n-1) \cdot 2+1=2 n-1 \end{aligned}$$

Thus, we have proven the statement.

## III. Solution.

Multiplying (1) by $a_{n}$:

$$a_{1} a_{n}+a_{2} a_{n}+\ldots+a_{n}^{2}=1$$

Write this for $n=1,2, \ldots, k$ and sum the resulting equations:

!

Since

$$\begin{gathered} \frac{1}{a_{k}^{2}}=\left(a_{1}+\ldots+a_{k}\right)^{2}=\sum_{n=1}^{k} a_{n}^{2}+2 \sum_{n<m} a_{n} a_{m}, \quad \text { it follows that } \\ \frac{1}{a_{k}^{2}} \geq \sum_{n=1}^{k} a_{n}^{2}+\sum_{n<m} a_{n} a_{m}=k \quad \text { and } \\ \frac{1}{a_{k}^{2}}=2\left(\sum_{n=1}^{k} a_{n}^{2}+\sum_{n<m} a_{n} a_{m}\right)-\sum_{n=1}^{k} a_{n}^{2}=2 k-\sum_{n=1}^{k} a_{n}^{2} \geq 2 k-a_{1}^{2}=2 k-1 \end{gathered}$$

Thus, $k \leq 1 / a_{k}^{2} \leq 2 k-1$, from which the statement follows immediately.