a[n+1]>a[n] a[n+2]^2=a[n+1]^2+2a[n+1]/a[n]+1/a[n]^2 >a[n+1]^2+2 (n>1) a[180]^2>a[4]^2+2*(180-4) =361 a[180]>19

fafafk wrote: a[n+1]>a[n] a[n+2]^2=a[n+1]^2+2a[n+1]/a[n]+1/a[n]^2 >a[n+1]^2+2 (n>1) a[180]^2>a[4]^2+2*(180-4) =361 a[180]>19 Can you explain how from a[n+2]^2=a[n+1]^2+2a[n+1]/a[n]+1/a[n]^2 >a[n+1]^2+2 (n>1) you came to a[180]^2>a[4]^2+2*(180-4) =361

Angelaangie wrote: fafafk wrote: a[n+1]>a[n] a[n+2]^2=a[n+1]^2+2a[n+1]/a[n]+1/a[n]^2 >a[n+1]^2+2 (n>1) a[180]^2>a[4]^2+2*(180-4) =361 a[180]>19 Can you explain how from a[n+2]^2=a[n+1]^2+2a[n+1]/a[n]+1/a[n]^2 >a[n+1]^2+2 (n>1) you came to a[180]^2>a[4]^2+2*(180-4) =361 because a(5)^2>a(4)^2+2 a(6)^2>a(5)^2+2 .............. a(180)^2>a(179)^2+2 add it all up .you can get it

Yup, all clear. Thanks a lot

We square the relation and we get the following: $$a_{n+2}^2 = a_{n+1}^2 + 2\frac{a_{n+1}}{a_n} + \left(\frac{1}{a_n}\right)^2 > a_{n+1}^2 + 2$$This holds since $a_{n+1} \geq a_n$ Telescoping this we get the following: $$a_{n+2}^2 > a_4^2 + 2(n+2-4)$$Since $a_4=3$, we have that $a_{180}^2 >9+2(180-4)=361 $ Since $a_n > 0$, for all $n$, we have that $a_{180} > 19$ . . .

A different (but long) solution: Claim (Key Claim): For any $t \in \mathbb Z_{>0}$, in the interval $[t,t+2]$, at most $2(t+1)$ terms of the sequence lie. Proof: Assume contrary that some for some $j$, all the terms $a_{j+1},a_{j+2},\ldots,a_{j + 2(t+1) + 1}$ lie in the interval $[t,t+2]$. Consider the sub-intervals \begin{align*} & \left[t ~,~ t + \frac{1}{t+1} \right), \left[t + \frac{1}{t+1} ~,~ t + \frac{2}{t+1} \right),\ldots,\left[ t + \frac{t}{t+1} ~,~ t+1 \right), \\ & \left[t+1 ~,~ t+1 + \frac{1}{t+2} \right), \left[ t+1 + \frac{1}{t+2} ~,~ t + 1 + \frac{2}{t+2} \right), \ldots, \left[ t+1 + \frac{t+1}{t+2} ~,~ t+2 \right] \end{align*}Note that each interval can contain at most one term of the sequence as $$a_{t + 2} = a_{t+1} + \frac{1}{a_t} \ge a_{t+1} + \frac{1}{a_{t+1}} > a_{t+1} + \frac{1}{ \lfloor a_{t+1} \rfloor + 1}$$Our assumption implies that exactly $1$ term of the sequence lies in each interval. But then, \begin{align*} 2 & \ge a_{j+2(t+1)+1} - a_{j+1} = \sum_{k=j+1}^{j + 2(t+1)} a_{k+1} - a_k = \sum_{k=j+1}^{j + 2(t+1)} \frac{1}{a_{k-1}} \\ & \ge \frac{1}{t} + \frac{1}{t + \frac{1}{t+1}} + \cdots + \frac{1}{t + \frac{t}{t+1}} + \frac{1}{t+1} + \frac{1}{t+1 + \frac{1}{t+2}} + \cdots + \frac{1}{t+1 + \frac{t}{t+2}} \\ &= \frac{1}{t} + \left( \frac{1}{t + \frac{1}{t+1}} + \cdots + \frac{1}{t + \frac{t}{t+1}} \right) + \left( \frac{1}{t+1 + \frac{1}{t+2}} + \cdots + \frac{1}{t+1 + \frac{t+1}{t+2}} \right) + \frac{1}{t+1} - \frac{1}{t+1 + \frac{t+1}{t+2}} \\ & > \frac{1}{t} + \frac{t}{t + \frac{t}{2}} + \frac{t+1}{t+1 + \frac{t+1}{2}} = \frac{1}{t} + \frac{2t}{2t+1} + \frac{2t+2}{2t+3} = 2 + \frac{1}{t} - \frac{1}{2t+1} - \frac{1}{2t + 3} > 2 \end{align*}So we obtain our desired contradiction. $\square$ Corollary: $a_j \ge t ~ \implies ~ a_{j + 2(t+1)} > t+2$. We return to the main problem now. Note that $a_3 = 2,a_4 = 3$. For any $l \in \mathbb Z_{> 0}$, let $$f(l) = 4 + \sum_{k=2}^{l+1} 4k$$Using our corollary and induction on $l$ yields $$a_{4 + f(l)} > 3 + 2l \qquad \forall ~ l \in \mathbb Z_{>0}$$Observe that $$f(8) = \sum_{k=2}^{9} 4k = 4 \cdot \sum_{k=2}^{9} k = 4 \cdot \left( \frac{9 \cdot 10}{2} - 1 \right) = 180 - 4 $$Hence, $$19 = 3 + 2(8) < a_{4 + f(8)} = a_{180}$$This completes the proof of the problem. $\blacksquare$

It's easy to see that: $$a_{n+2}^2 = a_{n+1}^2 + \frac{2a_{n+1}}{a_n }+ \frac{1}{a_n^2} > a_{n+1}^2 + 2.$$$$\implies a_{n+2}^2 > 2(n-2) + a_4^2 = 2n + 5$$Then $a_{180} > 19$

We claim that $a_n>\sqrt{2n+1}$ for all $n\ge4$, which we will show by induction. The base case follows since $a_3=2$, $a_4=\frac72$, and $\frac72>3$. For the induction step, note that $a_{n+1}>a_n$ for all $n\ge3$, from which it follows that $a_n>1$ for $n\ge3$. Let $f(x)=x+\frac1x$ which is continuous differentiable on $\mathbb R_{>1}$; on that domain we have $f'(x)=\frac{x^2-1}{x^2}>0$ so $f$ is strictly increasing. In addition, we establish the following inequality for $n\ge0$: $$\sqrt{2n+3}+\frac1{\sqrt{2n+3}}>\sqrt{2n+5}\Leftrightarrow2n+3+2+\frac1{2n+3}>2n+5\Leftrightarrow\frac1{2n+3}>0$$which is true. So if $a_{n+1}>\sqrt{2n+3}$ for some $n\ge3$, we have: $$a_{n+2}=a_{n+1}+\frac1{a_n}>a_{n+1}+\frac1{a_{n+1}}>\sqrt{2n+3}+\frac1{\sqrt{2n+3}}>\sqrt{2n+5}$$completing the induction. Finally, $a_{180}>19$.

Notice that \[a_{n+2}^2 = a_{n+1}^2 + \frac{2a_{n+1}}{a_n} + \frac 1{a_n^2} > a_{n+1}^2+ 2\]as $a_{n+1} > a_n$. So $a_{180}^2 > a_4^2 + 176 \cdot 2 = 9+352 = 361$ and thus $a_{180} > 19$.