IMO 2023 | PROBLEM 4 - SOLUTION - 9 July 2023

Problem 4.
Let $x_1,x_2,\dots,x_{2023}$ be pairwise different positive real numbers such that
$a_n=\sqrt{(x_1+x_2+\dots+x_n)\left(\frac{1}{x_1}+\frac{1}{x_2}+\dots+\frac{1}{x_n}\right)}$
is an integer for every $n=1,2,\dots,2023.$ Prove that $a_{2023} \geqslant 3034.$

proposed by Merlijn Staps, Netherlands

Solution 1

We claim $a_{n + 2} \ge a_n + 3$ which is clearly enough. Let $s_n = x_1 + x_2 + \dots + x_n$ and $h_n = \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}$ . Then

$a_{n + 1}^2 = s_{n + 1}h_{n + 1} = (s_n + x_{n + 1})\left(h_n + \frac{1}{x_{n + 1}}\right) = s_nh_n + \left(\frac{s_n}{x_{n + 1}} + x_nh_{n + 1}\right) + 1 = a_n^2 + \left(\frac{s_n}{x_{n + 1}} + x_{n + 1}h_n\right) + 1$
By AM-GM we have $\frac{s_n}{x_{n+1}} + x_{n+1}h_n \ge 2\sqrt{s_nh_n} = 2a_1$ and therefore $a_{n + 1}^2 \ge a_n^2 + 2a_n + 1 = (a_n + 1)^2$ . Thus $a_{n + 1} \ge a_n + 1$ for all $n$ , with equality if and only if $\frac{s_n}{x_{n+1}} = x_{n+1}h_n$ , or $x_{n+1}^2 = \frac{s_n}{h_n}$ . We finally prove that equality cannot happen twice in a row. Otherwise we have

$x_{n + 1}^2 = \frac{s_{n + 1}}{h_{n + 1}} = \frac{s_n + x_{n + 1}}{h_n + \frac{1}{x_{n + 1}}} = \frac{s_n + \sqrt{\frac{s_n}{h_n}}}{h_n + \sqrt{\frac{h_n}{s_n}}} = \frac{s_n}{h_n} = x_n^2$
A contradiction as $x_1, x_2, \dots, x_{2023}$ are all distinct.

Solution 2

Claim : $a_{n+2}>a_{n}+2$ which is equivalent to $a_{n+2} \ge a_{n}+3$ since $a_n$ is an integer. Hence $a_{2023} \ge a_1+\sum_{n=1}^{1011}3=3034$
Then prove the claim:

We define $X_n=x_1 + x_2 + \cdots + x_n$ and $Y_n=\frac{1}{x_1}+\cdots+\frac{1}{x_{n}}$ , so $a_n=\sqrt{X_n Y_n}$
$\begin{align*} a_{n+2}=&\sqrt{(x_1 + x_2 + \cdots + x_n+x_{n+1}+x_{n+2}) (\frac{1}{x_1}+\cdots+\frac{1}{x_{n}}+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &= \sqrt{(X_n+x_{n+1}+x_{n+2})(Y_n+\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &= \sqrt{a_n^2+X_n(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})+Y_n(x_{n+1}+x_{n+2})+(x_{n+1}+x_{n+2})(\frac{1}{x_{n+1}}+\frac{1}{x_{n+2}})} \\ &> \sqrt{a_n^2+ 4 \sqrt{X_nY_n}+4} \\ &= \sqrt{a_n^2+4a_n+4}=a_n+2 \end{align*}$

Solution 3

$a_{n+1}=\sqrt{(x_1+...+x_{n}+x_{n+1})(\frac{1}{x_1}+...+ \frac{1}{x_n}+ \frac{1}{x_{n+1}})}\Leftrightarrow$
$\Leftrightarrow a_{n+1}^2=a_{n+1}^2+x_{n+1}(\frac{1}{x_1}+...+ \frac{1}{x_n})+ \frac{1}{x_{n+1}}(x_1+...+x_n)+1\geq$
$\geq a_n^2 + 2\sqrt{(x_1+...+x_n)(\frac{1}{x_1}+...+\frac{1}{x_n})}+1=(a_n+1)^2\Leftrightarrow$
$\Leftrightarrow a_{n+1}\geq a_n +1$ ,

Using twice the above we get:
$a_{n+1}\geq a_{n-1}+2$

We will show by using twice when the equality holds in AM-GM, that the equality cannot hold. It should be true:

$x_n^2=\dfrac{x_1+...+x_{n-1}}{\frac{1}{x_1}+...+\frac{1}{x_{n-1}}}$

and:

$x_{n+1}^2=\dfrac{x_1+...+x_{n-1}+x_{n}}{\frac{1}{x_1}+... +\frac{1}{x_{n-1 }}+ \frac{1}{x_n}}= \dfrac{x_1+...+x_{n-1}+\sqrt{\dfrac{x_1+...+x_{n-1}}{\frac{1}{x_1}+...+ \frac{1}{x_{n-1 }}}}}{\frac{1}{x_1}+... +\frac{1}{x_{n-1 }}+ \sqrt{\dfrac{\frac{1}{x_1}+...+\frac{1}{x_{n-1}}}{x_1+...+x_{n-1}}}}=\dfrac{x_1+...+x_{n-1}}{\frac{1}{x_1}+...+\frac{1}{x_{n-1}}}=x_{n}^2$

which is inappropriate since the $x_n, x_{n+1$ are positive and different. So since the equality cannot be valid it is:

$a_{n+1}\geq a_{n-1}+3$

Using the above repeatedly we get:

$a_{2023}\geq a_1 + 3\cdot1011 =3$