Let $s_k = a_1 + \dots + a_k$, where $s_0 = 0$. By taking the reciprocal and simplifying, the inequality becomes \[ \frac{k}{a_{k+1}} \le a_k + \frac{k-1}{a_k}. \]Summing these all together, we now obtain a telescoping sum, and deduce the key fact \[ s_k \ge \frac{k}{a_{k+1}} \iff a_{k+1} \ge \frac{k}{s_k}. \]Now the result follows by induction: for any $n$, we have \[ s_{n+1} = s_{n} + a_{n+1} \ge s_n + \frac{n}{s_n} \ge n+1 \]where in the last step we use the fact that $s_n \ge n$.

Note that the condition rearranges into $\frac{a_{k+1}}{k}\geq\frac{a_k}{a_k^2+k-1}$ $\frac{k}{a_{k+1}}\leq a_k+\frac{k-1}{a_k}$ Summing from $k=1$ to $k=n$, we have that $a_1+a_2+\ldots+a_n\geq\frac{n}{a_{n+1}}$. Now, we proceed by induction, with the base case of $n=2$ following from $a_1+a_2\geq a_1+\frac{1}{a_1}\geq2$. For the inductive step, note that $a_1+a_2+\ldots+a_{n+1}\geq a_1+a_2+\ldots+a_n+\frac{n}{a_1+a_2+\ldots+a_n}\geq n+1$ since $a_1+a_2+\ldots+a_n\geq n$, as desired.

Write $b_k = \frac{1}{a_k}$. Now the condition is equivalent to $b_{k+1} \leq \frac{a_k^2 + (k-1)}{ka_k} = \frac{1}{k} \left( a_k + \frac{k-1}{a_k} \right) =\frac{1}{k} \left( \frac{1}{b_k} + (k-1)b_k \right)$. With this we observe that $b_2 \leq \frac{1}{b_1}$, $b_3 \leq \frac{1}{2} \left( \frac{1}{b_2} + b_2 \right) \leq \frac{1}{2} \left( \frac{1}{b_2} + \frac{1}{b_1} \right)$, $b_4 \leq \frac{1}{3} \left( \frac{1}{b_3} + 2b_3 \right) \leq \frac{1}{3} \left( \frac{1}{b_3} + \frac{1}{b_2} + \frac{1}{b_1} \right)$, and we can easily prove by induction that $b_{k+1} \leq \frac{1}{k} \left( \frac{1}{b_k} + \frac{1}{b_{k-1}} + \cdots + \frac{1}{b_1} \right)$ for all $k$. Now we go back to the $a_i$'s: the previous statement reads $\frac{1}{a_{k+1}} \leq \frac{1}{k} \left(a_k + a_{k-1} + \cdots + a_1 \right)$, or $a_{k+1} \geq \frac{k}{a_k+a_{k-1} + \cdots + a_1}$. Using this last inequality we can prove the desired result by induction as it has already been done in the previous posts.

Alternatively, once we get $a_k(a_1 + a_2 + \cdots + a_{k - 1}) \ge k$, summing from $k = 1$ to $k = n$ gives \[\sum_{1 \le i < j \le n}a_ia_j \ge \frac{n(n - 1)}{2}.\]It is easy to see that $a_1 + a_2 + \cdots + a_n \ge n$ from here.

We take notations from v_Enhance. Similarly get $s_ka_{k+1} \ge k$ for all $k \ge 1$. Now $$\frac{n-1}{2n}s^2_n \ge \frac{1}{2}((\sum_{i=1}^n a_i)^2-(\sum_ {i=1}^n a^2_i))=\sum_{i\not= j} a_ia_j=\sum_{i=1}^{n-1} s_ia_{i+1} \ge \frac{(n-1)n}{2} \implies s_n \ge n$$where the first inequality is just Cauchy-Schwarz. $\blacksquare$. Of course, equality is at $a_1=a_2= \cdots a_n=1$.

v_Enhance wrote: \[s_n + \frac{n}{s_n} \ge n+1 \] How is this true? Can someone explain?

mrowhed wrote: v_Enhance wrote: \[s_n + \frac{n}{s_n} \ge n+1 \] How is this true? Can someone explain? The function $f(x) = x + \frac{n}{x}$ is strictly increasing in the interval $[n,+\infty)$, and since $f(n) = n+1$, the assertion follows. To prove the former, just observe that if $y > x$ then $f(y) - f(x) = y-x + n \left(\frac{1}{y}- \frac{1}{x} \right) = y-x + n \left( \frac{x-y}{xy} \right) = (y-x) \cdot \left(1 - \frac{n}{xy} \right) > 0$. (In the last step we used that since $y > x \geq n$, we have $\frac{n}{xy} < \frac{n}{n^2} = \frac{1}{n} \leq 1$.) Alternatively, if you happen to know some basic calculus you can just check that the derivative $f'(x) = 1 - \frac{n}{x^2}$ is positive for $x \geq n$.

Another way: notice that for positive $x$, the inequality $x + \frac{n}{x} \geq n+1$ is equivalent to $x^2 + n \geq (n+1)x$, which is $x^2 - (n+1)x + n \geq 0$, and because the $LHS$ factors as $(x-1)(x-n)$ it is clear that this holds for $x \geq n$.

In fact, we can simply the problem by let $\begin{cases}m_{1} = a_{1} \\ m_{k + 1} = \dfrac{km_{k}}{m_{k}^{2} + k - 1}\end{cases}$ and show that $\sum_{i = 1}^{n}m_{i} \ge n$ since $a_{i} \ge m_{i} \quad \forall i \ge 1$ $\boxed{\textbf{Claim 1:}\; \frac{k}{m_{k + 1}} = \sum_{i = 1}^{k}m_{i} \quad \forall k \ge 1}$

Turn back to the problem, I'll show by induction. Easily, we have $m_{1} + m_{2} = m_{1} + \frac{1}{m_{1}} \ge 2$. Suppose the statement is true for $n \ge 2$, which means $\sum_{i = 1}^{n}m_{i} \ge n$. I'll show that it's also true for $n + 1$: The case $m_{n + 1} \ge 1$ is trivial since $\sum_{i = 1}^{n + 1}m_{i} = m_{1} + \sum_{i = 1}^{n}m_{i} \ge 1 + n$ The case $m_{n + 1} < 1$. I'll show that $\sum_{i = 1}^{n + 1}m_{i} \ge n + 1 \iff m_{n + 1} + \frac{n}{m_{n + 1}} \ge n + 1$. For convenient, setting $x = m_{n + 1}$: We have to show that $x + \frac{n}{x} \ge n + 1 \iff x^{2} - (n + 1)x + n \ge 0 \iff (x - 1)(x - n) \ge 0$ which is true since $x < 1$. Thus, the statement is true forall $n \ge 2$.

This is also India TST 2016, Day 1, Problem 2.

but it is not imo2015/#5

rightways wrote: but it is not imo2015/#5 He said original problem 5

My solution: By Mathematical Induction we know; $n=2\to a_1+a_2\geq \frac{1}{a_2}+a_2\geq 2$ where it is true by $AM-GM$ and in the condition if $k=1$ then $a_2\geq \frac{a_1}{a_1^2}\to a_1\geq \frac{1}{a_2}.$ Let $n=t\to a_1+...+a_t\geq t.$ Suppose that it is true, then we prove that $n=t+1.$ Then $a_1+a_2+...+a_t+a_{t+1}\geq t+1\to a_{t+1}\geq 1,$ where $a_1+...+a_t\geq t,$ for $t\geq 2.$ Assume that $a_{t+1}<1.$ Then $1>\frac{t\cdot a_t}{(a_t)^2+(t-1)}\to (a_t)^2-t\cdot a_t+(t-1)>0.$ We know if this inequality $>0,$ then $\Delta <0.$ Then $\Delta=t^2-4(t-1)=t^2-4t+4=(t-2)^2<0.$ But it isn't true and this is contradiction.

Ferid.---. wrote: My solution: By Mathematical Induction we know; $n=2\to a_1+a_2\geq \frac{1}{a_2}+a_2\geq 2$ where it is true by $AM-GM$ and in the condition if $k=1$ then $a_2\geq \frac{a_1}{a_1^2}\to a_1\geq \frac{1}{a_2}.$ Let $n=t\to a_1+...+a_t\geq t.$ Suppose that it is true, then we prove that $n=t+1.$ Then $a_1+a_2+...+a_t+a_{t+1}\geq t+1\to a_{t+1}\geq 1,$ where $a_1+...+a_t\geq t,$ for $t\geq 2.$ Assume that $a_{t+1}<1.$ Then $1>\frac{t\cdot a_t}{(a_t)^2+(t-1)}\to (a_t)^2-t\cdot a_t+(t-1)>0.$ We know if this inequality $>0,$ then $\Delta <0.$ Then $\Delta=t^2-4(t-1)=t^2-4t+4=(t-2)^2<0.$ But it isn't true and this is contradiction. Guys is this solution true? because if yes.problem is so easy.and what does original p5 mean?

The solution is very wrong... Just because a quadratic is more than zero for some value does not mean its discriminant is negative. This is only true if the quadratic is more than zero for EVERY value...

Define $s_n := a_1 + a_2 + \ldots + a_n$, for brevity. We prove by induction that $s_n \ge n$ for all $n \ge 2$. Note that $a_2 \ge \frac1{a_1}$ from letting $k=1$, so by AM-GM, $s_2 = a_1 + a_2 \ge a_1 + \frac1{a_1} \ge 2$. This establishes the base case $n=2$. Suppose that $s_m \ge m$ for some $m \ge 2$. Note that the given, $a_{k+1} \ge \dfrac{ka_k}{a_k^2+{k-1}}$, is equivalent to $\frac{k}{a_{k+1}} - \frac{k-1}{a_k} \le a_k$. Summing this inequality from $k=1$ to $k=m$ gives \[\frac{m}{a_{m+1}} \le a_1 + a_2 + \ldots + a_m = s_m\]so we can write \[s_{m+1} = s_m + a_{m+1} \ge s_m + \frac{m}{s_m} = f(s_m),\]where $f(x) := x + \frac mx$ for $x > 0$. It is easy to prove that the function $f$ is increasing on the interval $(\sqrt{m}, \infty)$, so \[s_{m+1} = f(s_m) \ge f(m) = m+1.\]This completes the induction and the proof.

CantonMathGuy wrote: It is easy to see that $a_1 + a_2 + \cdots + a_n \ge n$ from here. How did you get that?

Olenka wrote: CantonMathGuy wrote: It is easy to see that $a_1 + a_2 + \cdots + a_n \ge n$ from here. How did you get that? You can use the rearrangement inequality or more simply sum $a_i^2 + a_j^2 \ge 2a_ia_j$ for all pairs to obtain $\sum_{i=1}^n a_i^2 \ge \frac{2}{n-1} \sum_{1 \le i < j \le n} a_ia_j$. Then we have: $$ \sum_{i=1}^n a_i = \sum_{i=1}^n a_i^2 + \sum_{1 \le i < j \le n} 2a_ia_j \ge \frac{2n}{n-1} \sum_{1 \le i < j \le n} a_ia_j \ge \frac{2n}{n-1} \frac{n(n-1)}{2} = n^2 $$ This problem seemed very easy to me, and the proof naturally arrives if you follow v_enhance's line of reasoning. So does anyone have any idea how it was originally chosen as P#5 in IMO 2015? Did the jury have a tougher formulation, or were they trying to compensate for a relatively difficult day 1?

This problem was proposed by Dusan Djukic, Serbia. And it is almost the same with 2001 A5.

Define the sequence $b_k=\sum_{i=1}^ka_i$ for all $k\in \mathbb{Z}_{\geq 1}$. We claim that $a_k\geq \frac{k-1}{b_{k-1}}$ for all $k\in \mathbb{Z}_{\geq 2}$. To prove this, let’s proceed by induction: Base case: $k=2$ We see that $$a_2\geq \frac{1\cdot a_1}{a_1^2+(1-1)}=\frac{1}{a_1}=\frac{1}{b_1}=\frac{2-1}{b_{2-1}}$$ Inductive step: Suppose that $$a_l\geq \frac{k-1}{b_{k-1}}$$Then $$a_{l+1}\geq \frac{la_l}{a_l^2+l-1}\geq\frac{l\left(\frac{l-1}{b_{l-1}}\right)}{\left(\frac{l-1}{b_{l-1}}\right)^2+l-1}=\frac{l}{\left(\frac{l-1}{b_{l-1}}\right)}+(l-1)\div \left(\frac{l-1}{b_{l-1}}\right)\geq \frac{l}{a_l+b_{l-1}}=\frac{l}{b_1}$$as desired! We now show that $b_k\geq k$ for all $k\in \mathbb{Z}_{\geq 2}$, as the problem asked. Again, we proceed by induction. Base case: $k=2$ $$a_1+\frac{1}{a_1}\geq 2\sqrt{\left(\frac{1}{a_1}\right)(a_1)}=2$$ Inductive step: Suppose that $b_m\geq m$ Then $$b_{k+1}=b_k+a_{k+1}\geq b_k+\frac{k}{b_k}\geq k+\frac{k}{k}=k+1$$ So we are done! I think that if this problem had been P5 it would have been one of the P2s or P5s with a historically high number of solves, as in my opinion this problem is far easier than most P1s and P4s as all the ideas were rather natural (unless I missed something and my solution is completely wrong).

Flipping the inequality shows that \begin{align*} \frac{1}{a_{k+1}} &\leq \frac{a_k^2 + (k-1)}{ka_k} \\ \implies \frac{k}{a_{k+1}} &\leq a_k + \frac{k-1}{a_k} \\ \implies a_k &\geq \frac{k}{a_{k+1}} - \frac{k-1}{a_k} \\ \implies a_1 + a_2 + \dots + a_n &\geq \frac{n}{a_{n+1}} \end{align*}Now, we induct, with base case $n = 2$ obvious from AM-GM. Since $a_1 + a_2 + \dots + a_n \geq n$, then \[ a_1 + a_2 + \dots + a_n + a_{n+1} \geq (a_1 + a_2 + \dots + a_n) + \frac{n}{a_1 + a_2 + \dots + a_n} \geq n+1 \]because for some $0 < e < 1$, \begin{align*} e + \frac{n}{n+e} &> 1 \\ \iff en + e^2 + n &> n+e \\ \iff en + e^2 &> e \\ \iff n + e &> 1 \end{align*}which is obviously true. $\blacksquare$

(messed up initially ) First, we take the base case $n = 2$. Plugging $k = 1$ into the inequality yields: $$a_2 \geq \frac{1}{a_1} \implies a_1 + a_2 \geq a_1 + \frac{1}{a_1} \geq 2$$Now we take the reciprocal of the inequality and multiply by $k$ to get: $$\frac{k}{a_{k+1}} - \frac{k - 1}{k} \leq a_k$$Plugging in values of $k$ from $1$ to $k$ yields: $$\frac{1}{a_2} \leq a_1$$$$\frac{2}{a_3} - \frac{1}{a_2} \leq a_2$$$$\frac{3}{a_4} - \frac{2}{a_3} \leq a_3$$$$\dots$$$$\frac{k}{a_{k+1}} - \frac{k - 1}{a_k} \leq a_k$$Add all these inequalities to obtain $\frac{k}{a_{k+1}} \leq s_k$ where $s_k$ denotes $a_1 + \dots + a_k$. Assume the statement is true for $\{1, \dots, n\}$. $$s_{n+1} = s_n + a_{n + 1} \geq s_n + \frac{n}{s_n}$$We know $s_n \geq n \implies (s_n - 1)(s_n - n) \geq 0 \implies {s_n}^2 + n \geq s_n(n+1) \implies s_n + \frac{n}{s_n} \geq n+1$. Hence we are done.

Can't imagine this as a P5 It's very easy for that placement. First of all, assume that $k=1$ we have that. \[a_1+a_2 \geq a_1+\frac{1}{a_1} \geq 2\]Which follows by (AM-GM or rearrangement inequality). Now assume our problem holds for $k=n$ we will show fro $k=n+1$. Let's look back at the original statement. \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]If we take the reciprocals of both sides we have that. \[\frac{1}{a_{k+1}} \le \frac{a_k^2+(k-1)}{ka_k} \implies \frac{k}{a_{k+1}} \le \frac{a_k^2+(k-1)}{a_k} = a_k + \frac{k-1}{a_k}.\]So we have that \[\frac{k}{a_{k+1}} - \frac{k-1}{a_k}. \leq a_{k+1}\]Which means that. \[\sum_{k=1}^{n+1} \frac{k}{a_{k+1}} - \frac{k-1}{a_k} = \frac{n}{a_{n+1}} \]So we have that. \[\frac{n}{a_{n+1}} \le a_1 + a_2 + \ldots + a_n \implies a_{n+1} \geq \frac{n}{\sum_{k=1}^{n+1}a_k}\]But this means that by our inductive hypothesis ($a_1+a_2+.....+a_n \geq n$) , we have that. \[a_1+a_2+....a_{n+1} \geq a_1+a_2+\cdots+a_n+\frac{n}{a_1+a_2+\cdots+a_n} = \frac{(a_1+a_2+\ldots +a_n)^2+n}{a_1+a_2+\ldots +a_n} \]\[ \geq \frac{(n+1)(a_1+a_2+\ldots +a_n)}{a_1+a_2+\ldots + a_n}=n+1 \]In which the last inequality ($\frac{(a_1+a_2+\ldots +a_n)^2+n}{a_1+a_2+\ldots +a_n} \geq \frac{(n+1)(a_1+a_2+\ldots +a_n)}{a_1+a_2+\ldots + a_n}$) holds by rearrangement inequality.

A little one before the hay is hit. We prove via induction. Rearranging the initial equation, we derive that \begin{align*} a_k + \frac{k - 1}{a_k} &\geq \frac{k}{a_{k + 1}} \\ \implies a_k + a_{k - 1} + \cdots + a_2 + a _1 &\geq \left(\frac{k}{a_{k + 1}} - \frac{k - 1}{a_k} \right) + \left(\frac{k - 1}{a_k} + \frac{k - 2}{a_{k - 1}} \right) + \cdots + \left(\frac{2}{a_3} - \frac{1}{a_2} \right) + \frac{1}{a_2} \\ \implies s_k &\geq \frac{k}{a_{k + 1}}. \end{align*}Rearranging the final expression also yields $a_{k + 1} \geq k/s_k$. Assume the result holds true for all integers at most $k$. From this we find by our inductive hypothesis that \[s_{k + 1} = s_k + a_{k + 1} \geq s_k + \frac{k}{s_k} \geq k + 1, \]as $s_k \geq k$, which was to be shown. $\blacksquare$

oops reciprocating hard Let $x_n = a_1 + a_2 + \cdots + a_n$. We prove $x_n \ge n$ with induction on $n$. For $n = 2$, we note that $a_2 \ge \tfrac{1}{a_1}$, so $a_1 + a_2 \ge a_1 + \tfrac{1}{a_1} \ge 2$ by AM-GM. Now, assume the statement holds for $n$; we will show it holds for $n + 1$. Note that by reciprocating the given equation, we can rearrange it as $$\frac{k}{a_{k + 1}} \le a_k + \frac{k - 1}{a_k} \implies a_k \ge \frac{k}{a_{k + 1}} - \frac{k - 1}{a_k}.$$Adding this up from $k = n$ down to $k = 1$, we have $x_n \ge \tfrac{n}{a_{n + 1}}$, or $a_{n + 1} \ge \tfrac{n}{x_n}$. We wish to show that $$x_n + \frac{n}{x_n} \ge n + 1 \iff x_n^2 - (n + 1)x_n + n \ge 0 \iff (x_n - 1)(x_n - n) \ge 0,$$which holds since $x_n \ge n$ by the inductive hypothesis.

\[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\]\[\implies\frac{1}{a_{k+1}}\leq\frac{a_k^2+(k-1)}{ka_k}\implies\frac{k}{a_{k+1}}\leq a_k+\frac{k-1}{a_k}\]\[\implies a_k\geq \frac{k}{a_{k+1}}-\frac{k-1}{a_k}\]\[\implies \sum_{i=1}^{n}a_i\geq\frac{n}{a_{n+1}}\implies a_{n+1}\geq\frac{n}{S(n)}\cdots(i)\]Let $S(n)=a_1+a_2+\cdots+a_n$ We take help of induction: Base case: Plugging in $k=1$ in the original inequality, we get: $a_2\geq\frac{1}{a_1} \implies a_1+a_2\geq a_1+\frac{1}{a_1} \geq 2$ (by AM-GM). $\implies \boxed{S(2)\geq 2}$ Assuming that $S(k)\geq k$, Inductive Step: $$S(k)+a_{k+1} \geq S(k)+\frac{k}{S(k)}\cdots\text{from $(i)$}$$$$\implies S(k+1) \geq S(k)+\frac{k}{S(k)}$$ $$S(k)-k\geq 0, S(k)-1 \geq 0\implies (S(k)-k)(S(k)-1)\geq 0 \implies S(k)+\frac{k}{S(k)}\geq k+1$$$$\implies \boxed{S(k+1)\geq k+1}$$Therefore, the claim is also true for $n=k+1$ and by the principle of mathematical induction, we're done!

Note that $\frac1{a_{k+1}}\le \frac{a_k}{k}+\frac{k-1}{ka_k}$. This means that $$\frac1{a_2}\le a_1$$and $$\frac1{a_3}\le \frac{a_2}{2}+\frac{1}{2a_2}\le \frac{a_1+a_2}{2}$$and $$\frac1{a_4}\le \frac{a_3}{3}+\frac{2}{3a_3}\le \frac{a_1+a_2+a_3}{3}$$and that $$\frac{1}{a_{k+1}}\le \frac{a_1+\dots+a_k}{k}$$in general, which is easy to prove by induction. Since $\frac1{a_2}\le a_1$, $a_1+a_2\ge 2\sqrt{a_1a_2}\ge 2$, as desired. Assume that $a_1+\dots+a_k\ge k$ for some $k$. Since $\frac1{a_{k+1}}\le \frac{a_1+\dots+a_k}{k}$, $$a_1+\dots+a_{k+1}\ge a_1+\dots+a_k+\frac{k}{a_1+\dots+a_k}\ge k+1,$$which finishes our induction, as desired.

Nice problem. We proceed with induction, the base case of $n = 2$ follows as $a_1 + a_2 = a_1 + \dfrac{1}{a_1} \ge 2$ by AM-GM. For the inductive step assume the claim is true for some $n \ge 2$ and we want to prove for $n + 1$. Note that $a_{n + 1} \ge 1$ immediately implies the desired conclusion so allow $a_{n + 1} < 1$. The given condition can be manipulated to yield \[a_k \ge \dfrac{k}{a_{k + 1}} - \dfrac{k - 1}{a_k} \implies a_1 + a_2 + \dots + a_{n + 1} \ge a_{n + 1} + \dfrac{n}{a_{n + 1}} \]but \[(a_{n + 1} - 1)(a_{n + 1} - (n + 1)) \ge 0 \implies a_{n + 1} + \dfrac{n}{a_{n + 1}} \ge n + 1\]and we are done.

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just posting for storage

Note that the condition resolves to $\frac{k}{a_{k + 1}} \le \frac{k - 1}{a_k} + a_k$, so we can telescope to get $\frac{k}{a_{k + 1}} \le a_1 + \cdots a_k$. We prove the statement inductively for $n = k + 1$ for all $k \ge 1$. Base Case: We get $a_2 \ge \frac{1}{a_1}$, so we are done by AM-GM. Now the for the inductive step, we are guaranteed the sum on the right is at least $k$. If it is at least $k + 1$, we are done. Otherwise, let the sum be $x$, then we desire $\frac kx + x \ge k + 1$ for all $x \in [k, k+ 1)$, which is obvious by noting the statement is true at $x = k$ and noting that the first derivative of $\frac kx + x$ is $1 - \frac{k}{x^2}$ is positive for all $x$ in the desired interval.

How is this IMO P5? (see attached.) iirc IMO 2015/5 is a FE Let $b_k=\frac1{a_k}$ for all $k$. Then the given condition rearranges to the inequality \[(k-1)b_k\le\frac1{b_{k-1}}+(k-2)b_{k-1}\]for all $k\ge2$. In particular, note that $\frac1{b_1}+\frac1{b_2}\ge\frac1{b_1}+b_1\ge2$ for $n=2$. By induction, suppose that $\sum_{i=1}^{n-1}\frac1{b_i}\ge n-1$. If $b_n<1$, we are done. Otherwise, summing our given condition from $k=2$ to $k=n-1$ and adding $\frac1{b_n}$ implies that $\sum_{i=1}^n\frac1{b_i}\ge(n-1)b_n+\frac1{b_n}\ge n$ since $b_n\ge1$. $\blacksquare$

Just for the sake of storage: Let us strong induct in $n$ Base case: Taking $k=1$ gives us $$a_2 \geq \frac{a_1}{a_1^2}=\frac{1}{a_1} \Rightarrow a_1+a_2 \geq a_1 +\frac{1}{a_1} \geq 2$$Induction Hypothesis: Suppose that $\sum_{i=1}^n a_i \geq n$ for all $n \leq t$ Inductive Step: Let´s consider two cases: 1) If $a_t > t-1$, since $\sum_{i=1}^{t-1} a_i \geq t-1$, we get that $\sum_{i=1}^{t+1} a_i \geq \sum_{i=1}^t a_i > 2(t-1) \geq t+1$ $\blacksquare$ 2)If $a_t \leq t-1$. Note that $$a_{t+1} \geq \frac{ta_t}{a_t^2+(t-1)} \geq$$$$\iff ta_t \geq a_t^2 +(t-1)$$$$\iff (t-2)(a_t-1 \geq a_t -2a_t +1=(a_t-1)^2$$$$\iff t-2 \geq a_t-1 \iff t-1\geq a_t$$So $\sum_{i=1}^{t}a_i + a_{t+1} \geq t+1$ $\blacksquare$ Obs: Note that we could divide $(a_t-1)$ because if $a_t=1$, the result we follow immediately.

Note that the inequality condition is equivalent to $$a_k \geq \frac{k}{a_{k+1}} - \frac{k-1}{a_k}.$$Therefore adding everything for $k=1, 2, \cdots, m$ yields $$a_{m+1} \geq \frac{m}{s_m}$$for all positive integers $m.$ Now, we proceed with induction. The base case, $n=2,$ is trivial as $a_1+a_2 \geq a_1+\frac{1}{a_1} \geq 2$ by AM-GM. Now, if the proposition holds for $n=k \geq 2$ observe that $$s_{k+1}=a_{k+1}+s_k \geq \frac{k}{s_k}+s_k.$$Now $$\frac{k}{s_k}+s_k \geq k+1 \iff (s_k-1)(s_k-k) \geq 2$$which is clearly true by the inductive hypothesis, so the proposition holds for $n=k+1.$ Thus our induction is complete, and we are done. QED