amparvardi wrote: Find all sets of real numbers $\{a_1,a_2,\ldots, _{1375}\}$ such that \[2 \left( \sqrt{a_n - (n-1)}\right) \geq a_{n+1} - (n-1), \quad \forall n \in \{1,2,\ldots,1374\},\] and \[2 \left( \sqrt{a_{1375}-1374} \right) \geq a_1 +1.\] Put $x_n:=a_n-(n-1)$. Then we have $x_n-x_{n+1}\ge(\sqrt{x_n}-1)^2 \quad \forall n \in \{1,2,\ldots,N-1\},$ $x_N-x_{1}\ge(\sqrt{x_N}-1)^2$ Adding them all up yields $0\ge\sum_{n=1}^N(\sqrt{x_n}-1)^2$. So for all $n$ we have $x_n=1$, i.e. $a_n=n$.

define $a_n-(n-1)=b_n \Longrightarrow 2\sqrt{b_n}\ge b_{n+1}+1$ and $2\sqrt{b_{1375}}\ge b_1+1$ and it is obvious that $b_i>0$ by AM-GM $b_n+1\ge 2\sqrt{b_n}\ge b_{n+1}+1 \Longrightarrow b_n\ge b_{n+1}$ and $b_{1375}\ge b_1$ $\Longrightarrow b_1\ge b_2\ge...\ge b_{1375}\ge b_1 \Longrightarrow b_1=b_2=...b_{1375}=1 \Longrightarrow a_i=i$

it's interesting! see the time of posts!