suppose $a_n < \frac{1}{n}$ for some n. $a_{n-1} = a_n^2 + a_n < \frac{1}{n^2}+\frac{1}{n} < \frac{1}{n-1}$ apply this ineq recursively $a_n < \frac{1}{n} \implies \cdots \implies a_1 < 1$ (=><=) this soln differs from post #2 in that it avoids the use of the quadratic formula.

Direct induction applies, it is suffices to show that $a_{n+1}\geq \frac{1}{n+1}$ given that \[a_{n+1}^2+a_{n+1}=a_n\geq \frac{1}{n}\]which follows from \[\left(a_{n+1}+\frac{1}{2}\right)^2\geq \frac{1}{n}+\frac{1}{4}\]and since $a_{n+1}$ is positive, we have \[a_{n+1}\geq \sqrt{\frac{1}{n}+\frac{1}{4}}-\frac{1}{2}\geq \frac{1}{n+1}\]since expanding it gives us \[\frac{1}{n}\geq \frac{1}{(n+1)^2}+\frac{1}{n+1} \iff 1\geq 0.\]

Use induction. The case $n=1$ is true. Now assume that for a $k$ we have $a_k \ge \frac{1}{k}$. We want to prove that $a_{k+1} \ge \frac{1}{k+1}$. For the sake of contradiction, assume $a_{k+1} < \frac{1}{k+1}$. Then, \[\frac{1}{k} \le a_k=a_{k+1}^2+a_{k+1} <\frac{1}{(k+1)^2}+\frac{1}{k+1}=\frac{k+2}{(k+1)^2}.\]Cross multiplying, \[1 < \frac{k^2+2k}{(k+1)^2},\]which is false. Hence, $a_{k+1} \ge \frac{1}{k+1}$ and we are finished by induction.

Using the quadratic formula and noting $a_{n+1}>0,$ we see $$a_{n+1}=\frac{\sqrt{4a_n+1}-1}{2}.$$We proceed by induction. Our base case $n=1$ is evident. If $a_n>1/n,$ we claim $a_{n+1}>1/(n+1).$ Indeed, since $n$ is positive, \begin{align*}1\ge 0&\implies (n+1)^2\ge n(n+1)+n\\&\implies \frac{4}{n}+1\ge\frac{4}{n+1}+\frac{4}{(n+1)^2}+1\\&\implies\frac{4}{n}+1\ge\left(1+\frac{2}{n+1}\right)^2\\&\implies(n+1)\sqrt{\frac{4}{n}+1}\ge n+3\\&\implies a_{n+1}=\frac{\sqrt{\frac{4}{n}+1}-1}{2}\ge\frac{1}{n+1}.\end{align*}Hence, our induction is complete. $\square$

One-liner: $$(\frac{1}{n+1})^2+\frac{1}{n+1}=\frac{1}{n+1}(1+\frac{1}{n+1})=\frac{n+2}{(n+1)^2}<\frac{n+2+\frac{1}{n}}{(n+1)^2}=\frac{1}{n} (\text{induction})$$

parmenides51 wrote: Let $a_1, a_2, a_3, . . .$ be a sequence of positive real numbers that satisfies $a_1 = 1$ and $a^2_{n+1} + a_{n+1} = a_n$ for every natural number $n$. Prove that $a_n \ge \frac{1}{n}$ for every natural number $n$. Who is the author?