Team Selection Test for EGMO 2024

Rearranging the condition, we obtain $(a_n-a_{n-1}{)}^2\ge 2a_{n-1}{a}_{n-2}$ and since the sequence is increasing we have $a_n\ge a_{n-1}+\sqrt{2a_{n-1}{a}_{n-2}}$. The inequality will be proved by induction over $n$.

For the base cases $n=2,3$ using the monotonicity we have $a_2\ge a_1$ and $a_3\ge a_2+\sqrt{2a_2{a}_1}>a_2+a_1$.

Assume that the assertion is true for $2,\dots ,n-1$. There are two possible cases:

If $a_{n-1}<2a_{n-2}$, then we have $a_n\ge a_{n-1}+\sqrt{2a_{n-1}{a}_{n-2}}\ge 2a_{n-1}$ and from the induction hypothesis we obtain $a_n\ge 2a_{n-1}\ge$ $$a_{n-1}+a_{n-2}+\cdots +a_1.$$ If $a_{n-1}\ge 2a_{n-2}$, then we have $a_n\ge a_{n-1}+\sqrt{2a_{n-1}{a}_{n-2}}\ge a_{n-1}+2a_{n-2}$ and from the induction hypothesis we obtain $a_n\ge a_{n-1}+2a_{n-2}\ge a_{n-1}+a_{n-2}+\cdots +a_1$.

We are done.

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Alternative solution.

We will again use induction. Cases $n=2,3$ are covered in Solution 1. Assume that the assertion is true for $n=1,2,\dots ,k$, we will prove that it is true for $n=k+1$. On the contrary, assume that $a_{k+1}<a_1+\cdots +a_k$. Then, we have $a_1+\cdots +a_{k-1}>a_{k+1}-a_k$. Since both sides are positive, we can take squares and use the problem condition to get $$(a_1+\cdots +a_{k-1}{)}^2>(a_{k+1}-a_k{)}^2\ge 2a_k{a}_{k-1}$$ From the inductive hypothesis we know that $a_k\ge a_1+\cdots +a_{k-1}$. Combining last two inequalities we get $$a_1+\cdots +a_{k-1}>2a_{k-1}\text{ or }{a}_1+\cdots +a_{k-2}>a_{k-1},$$ which contradicts to the inductive hypothesis. Thus, $$a_{k+1}\ge a_1+\cdots +a_k.$$