International Mathematics Competition
for University Students
2026

IMC2026: Day 2, Problem 9

Problem 9. Let \(\displaystyle a_1,a_2,\ldots\) be an infinite sequence of positive real numbers satisfying

\(\displaystyle a_1 + a_2 + \cdots + a_{2n-1} = a_n^2 \)

for all positive integers \(\displaystyle n\). Prove that \(\displaystyle a_n \geq 2n-1\) for all positive integers \(\displaystyle n\).

Ilya I. Bogdanov, MIPT, Moscow and Aleksandr Kuznetsov, SPbU, Saint Petersburg

Solution 1. The condition applied to \(\displaystyle n=1\) gives \(\displaystyle a_1=1\), so the desired inequality holds for \(\displaystyle n=1\).

Denote \(\displaystyle \delta_n=a_{n+1}-a_n\). Say that an index \(\displaystyle i\) is regular if \(\displaystyle a_{i+1}\geq 2i+1\); otherwise \(\displaystyle i\) is irregular. Our aim is to prove that all indices are regular.

Say that our sequence is \(\displaystyle \mu\)-good if \(\displaystyle \delta_i\geq\mu\) for all irregular indices \(\displaystyle i\). We will show that our sequence is \(\displaystyle \mu\)-good for all \(\displaystyle \mu<2\). This yields the desired inequality; indeed, otherwise, choosing the minimum irregular index \(\displaystyle n\), we get \(\displaystyle a_{n+1}-a_n<(2n+1)-(2n-1)=2\), so the sequence would not be \(\displaystyle \mu\)-good for some \(\displaystyle \mu<2\).

Notice that \(\displaystyle a_{n+1}^2-a_n^2=a_{2n}+a_{2n+1}>0\), so \(\displaystyle a_{n+1}>a_n\) for all \(\displaystyle n\), and hence the sequence is \(\displaystyle 0\)-good. Now the desired statement follows from the Claim below.

Claim. If our sequence \(\displaystyle (a_n)\) is \(\displaystyle \mu\)-good for some \(\displaystyle 0\leq \mu<2\), then it is also \(\displaystyle \nu\)-good for \(\displaystyle \nu=\frac\mu2+1\).

Proof. Consider any irregular index \(\displaystyle n\). Notice that \(\displaystyle a_k-a_{n+1}\geq (k-n-1)\mu\) for every \(\displaystyle k>n+1\); indeed, if all indices \(\displaystyle n+1,n+2,\dots,k-1\) are irregular, then this follows from \(\displaystyle (a_n)\) being \(\displaystyle \mu\)-good. Otherwise, let \(\displaystyle i\) be the maximum regular index not exceeding \(\displaystyle k-1\). Then

\(\displaystyle a_k-a_{n+1}=(a_k-a_{i+1})+(a_{i+1}-a_{n+1})\geq \mu(k-i-1)+((2i+1)-(2n+1))=\mu(k-i-1)+2(i-n)>\mu(k-n-1), \)

as desired.

Therefore,

\(\displaystyle \delta_n(2a_{n+1}-\delta_n)=a_{n+1}^2-a_n^2 =a_{2n}+a_{2n+1}\geq 2a_{n+1}+(2n-1)\mu, \)

This inequality easily yields \(\displaystyle \delta_n>1\), so in particular \(\displaystyle a_{n+1}>2\). Then the function \(\displaystyle f(x)=x(2a_{n+1}-x)\) increases on \(\displaystyle x\in[0,2]\), and in order to prove \(\displaystyle \delta_n\geq\nu\) it suffices to show that \(\displaystyle f(\nu)\leq 2a_{n+1}+(2n-1)\mu\). This inequality rewrites as

\(\displaystyle \mu a_{n+1}-\left(\frac \mu2+1\right)^2\leq (2n-1)\mu \iff \mu(a_{n+1}-2n-1)\leq \left(\frac\mu2-1\right)^2. \)

This last inequality holds, since the left hand part is negative, while the right hand part is positive.

Solution 2. First, let us prove that the required inequality holds asymptotically, namely Lemma. \(\displaystyle \liminf\limits_{n\to\infty} \frac{a_n}{2n-1} \geq 1\)

Proof. Let \(\displaystyle A\) denote this limit inferior. Clearly, the sequence is monotonically increasing. Then \(\displaystyle a_n^2 = a_1+a_2+\dots+a_{2n-1}\geq na_n\), hence \(\displaystyle A\geq \frac{1}{2}>0\).

Let \(\displaystyle \varepsilon>0\). We know that \(\displaystyle a_n\geq (A-\varepsilon)(2n-1)\) for all sufficiently large \(\displaystyle n\). Then, from the inequality

\(\displaystyle a_n^2 = a_1+a_2+\dots+a_{2n-1}\geq (A-\varepsilon)(1+3+\ldots+4n-3) +O(1)= (A-\varepsilon)(2n-1)^2+O(1) \)

we obtain \(\displaystyle A^2\geq A-\varepsilon\). Letting \(\displaystyle \varepsilon\) tend to zero, we get \(\displaystyle A\geq 1\).

Let us make the substitution \(\displaystyle b_n=a_n-(2n-1)\). The recurrence relation takes the form

\(\displaystyle b_n(b_n+2(2n-1)) = b_1+\ldots+b_{2n-1}, \)

and the lemma implies that \(\displaystyle \liminf\limits_{n\to\infty} \frac{b_n}{2n-1} \geq 0\). Consider \(\displaystyle B = \inf\limits_{n}\frac{b_n}{2n-1}\). If \(\displaystyle B<0\), then the infimum is attained at some \(\displaystyle k\in \mathbb N\), i.e. \(\displaystyle b_k=B(2k-1)\) (otherwise we get a contradiction with the lemma). Then

\(\displaystyle B(B+2)(2k-1)^2 = b_k(b_k+2(2k-1)) = b_1+\ldots+b_{2k-1}> B(1+3+\ldots+4k-3) = B(2k-1)^2. \)

This yields \(\displaystyle B< -1\). On the other hand, \(\displaystyle a_n> 0\) implies \(\displaystyle B> -1\) and we get a contradiction.

Solution. [2 (by Dan Carmon)] Begin with observing \(\displaystyle a_1 = a_1^2\), so \(\displaystyle a_1 = 1\) from positivity. It follows that \(\displaystyle a_n^2 = \sum_{i=1}^{2n-1} a_i \ge a_1 = 1\) so \(\displaystyle a_n \ge 1 = (2n-1)^0\) for every \(\displaystyle n\). Applying the same technique again gives \(\displaystyle a_n^2 = \sum_{i=1}^{2n-1} a_i \ge 2n-1\) so \(\displaystyle a_n \ge \sqrt{2n-1} = (2n-1)^{1/2}\). Applying this technique again and again, we will prove the following claim by induction on \(\displaystyle m\): Let \(\displaystyle m \ge 0\), and set \(\displaystyle c_m = 1 - 2^{-m}\). Then there is a constant \(\displaystyle C_m > 0\) such that \(\displaystyle a_n \ge C_m (2n-1)^{c_m}\) for all \(\displaystyle n \ge 1\). Moreover, we will get a recurrence relation for \(\displaystyle C_m\), and show that \(\displaystyle \lim_{m \to \infty} C_m = 1\). It would follow that \(\displaystyle a_n \ge \lim_{m \to \infty} C_m (2n-1)^{c_m} = 2n-1\), as we are asked to show. Proof. We have already established the \(\displaystyle m=0, 1\) (\(\displaystyle c_0 = 0\), \(\displaystyle c_1 = 1/2\)) with \(\displaystyle C_0 = C_1 = 1\). Let \(\displaystyle m\ge 1\) and suppose \(\displaystyle a_n \ge C_m (2n-1)^{c_m}\) for every \(\displaystyle n\). Write \(\displaystyle c = c_m\) and \(\displaystyle C = C_m\) for brevity. Since \(\displaystyle c < 1\), the function \(\displaystyle x^c\) is concave, and therefore have \(\displaystyle x^c \ge \frac12 \int_{x-1}^{x+1} t^c dt\). Thus

$$\begin{align*} a_n^2 &= \sum_{k=1}^{2n-1} a_k \ge C \sum_{k=1}^{2n-1} (2k-1)^c \ge \frac{C}{2} \sum_{k=1}^{2n-1} \int_{2k-2}^{2k} t^c dt = \frac{C}{2} \int_{0}^{4n-2} t^c dt = \frac{C}{2} \frac{(4n-2)^{c+1}}{c+1} \\ & = \frac{2^c C}{1+c} (2n-1)^{1+c}. \end{align*}$$

Recall \(\displaystyle c = c_m = 1-2^{-m}\) hence \(\displaystyle \frac{1+c}{2} = c_{m+1}\). Taking the square root of the above inequality yields

\(\displaystyle a_n \ge \sqrt{\frac{2^{-2^{-m}}}{1 - 2^{-m-1}} C_m} \cdot (2n-1)^{c_{m+1}} \)

emacs p9 Which completes the inductions step for \(\displaystyle C_{m+1} = \sqrt{B_m C_m}\), where \(\displaystyle B_m = \frac{2^{-2^{-m}}}{1 - 2^{-m-1}}\).

Observe that \(\displaystyle B_m \to 1\) as \(\displaystyle m \to \infty\) (since \(\displaystyle 2^{-m} \to 0\)), and the sequence \(\displaystyle C_m\) is obtained by repeatedly averaging (geometrically) the previous term with the terms of \(\displaystyle B_m\). It is well known (a standard exercise in calculus 1) that this implies \(\displaystyle C_m\) also converges and to the same limit as \(\displaystyle B_m\), as we claimed. This can be shown directly by limits calculus; another method is to apply Cesaro's theorem on geometric means to the sequence \(\displaystyle D_k\) defined by \(\displaystyle D_1 = C_1\), \(\displaystyle D_k = B_{\lceil \log_2(k) \rceil}\) (i.e., the first elements are \(\displaystyle C_1, B_1, B_2, B_2, B_3, B_3, B_3, B_3, B_4, \dots\)), which clearly has the same limit as \(\displaystyle B_m\), and each \(\displaystyle C_m\) is just the geometric mean of the first \(\displaystyle 2^{m-1}\) elements of \(\displaystyle D_k\).