(a) We have $a_{n+1} + n \leq 2\sqrt{a_n\cdot\left(n+1\right)}$ by our problem hypothesis. On the other hand, the AM-GM inequality yields $2\sqrt{a_n\cdot\left(n+1\right)} \leq a_n + \left(n+1\right)$. Thus, $a_{n+1} + n \leq a_n + \left(n+1\right)$, so that $a_{n+1} - a_n \leq 1$. Similarly, $a_{n+2} - a_{n+1} \leq 1$, $a_{n+3} - a_{n+2} \leq 1$, ..., $a_{2n-1} - a_{2n-2} \leq 1$, $a_{2n} - a_{2n-1} \leq 1$. Adding these inequalities together, we get $a_{2n} - a_n$ $= \left(a_{2n} - a_{2n-1}\right) + \left(a_{2n-1} - a_{2n-2}\right) + ... + \left(a_{n+3} - a_{n+2}\right) + \left(a_{n+2} - a_{n+1}\right) + \left(a_{n+1} - a_n\right)$ $\leq 1 + 1 + ... + 1 + 1 + 1 = n$, so that $a_n - a_{2n} \geq -n$. Adding this to $a_n + a_{2n} \geq 3n$, we get $2a_n \geq 2n$, so that $a_n \geq n$, what proves (a). (b) Just take $a_n = n+1$ for every n. The condition $a_n + a_{2n} \geq 3n$ is clearly satisfied ($\left(n+1\right)+\left(2n+1\right) = 3n+2 \geq 3n$), and the condition $a_{n+1} + n \leq 2\sqrt{a_n\cdot\left(n+1\right)}$ is also satisfied (even the $\leq$ sign can be replaced by the = sign, since $a_{n+1} + n = \left(n+2\right) + n = 2\left(n+1\right) = 2\sqrt{\left(n+1\right)^2}$ $= 2\sqrt{\left(n+1\right)\cdot\left(n+1\right)} = 2\sqrt{a_n\cdot\left(n+1\right)}$ ). Darij

I have an alternate proof of (a). Assume that, for the sake of contradiction, that for some $n$, $a_n<n$. Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$. Thus, $a_{n+1}<n+1$. Repeating this $n-1$ times, $a_{2n}<2n$. Thus, $a_{2n}+a_n<3n$, contradiction. And so $a_n\ge n$ for all positive integers $n$.

GoldenFrog1618 wrote: Assume that, for the sake of contradiction, that for some $n$, $a_n<n$. Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$. Thus, $a_{n+1}<n+1$. Repeating this $n-1$ times, $a_{2n}<2n$. Thus, $a_{2n}+a_n<3n$, contradiction. And so $a_n\ge n$ for all positive integers $n$. This proof is wrong. Because you have $a_n<n$ for some n, and $a_{n+1}<n+1$ for some n only, too. So we cannot "repeat" them for other n.

TT-P wrote: GoldenFrog1618 wrote: Assume that, for the sake of contradiction, that for some $n$, $a_n<n$. Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$. Thus, $a_{n+1}<n+1$. Repeating this $n-1$ times, $a_{2n}<2n$. Thus, $a_{2n}+a_n<3n$, contradiction. And so $a_n\ge n$ for all positive integers $n$. This proof is wrong. Because you have $a_n<n$ for some n, and $a_{n+1}<n+1$ for some n only, too. So we cannot "repeat" them for other n. No I don't think so.........Because we can assume n + 1 = A ....and it is right with A+1 so on

mathkhtn wrote: TT-P wrote: GoldenFrog1618 wrote: Assume that, for the sake of contradiction, that for some $n$, $a_n<n$. Thus, $a_{n+1}+n<2\sqrt{a_n(n+1)}<2\sqrt{n^2+n}<2\sqrt{n^2+n+\frac{1}{4}}<2n+1$. Thus, $a_{n+1}<n+1$. Repeating this $n-1$ times, $a_{2n}<2n$. Thus, $a_{2n}+a_n<3n$, contradiction. And so $a_n\ge n$ for all positive integers $n$. This proof is wrong. Because you have $a_n<n$ for some n, and $a_{n+1}<n+1$ for some n only, too. So we cannot "repeat" them for other n. No I don't think so.........Because we can assume n + 1 = A ....and it is right with A+1 so on No, your proof is not true, sorry.