# Given $a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$, prove $S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2$

Suppose a sequence of positive real numbers with $$a_{k+1} \ge \frac{k a_{k}}{(a_{k}^{2} + k-1)}, \:\: k > 0$$ prove that $$S_{n} = a_{1} + .. + a_{n} \ge n, \:\: n \ge 2$$

Solution: I will show two different approaches, one is finished, the other one is still confusing

• Finished approach:

By induction, for the base case we have $$a_{2} \ge a_{1}/a_{1}^{2} = 1/a_{1}^ \implies a_{1} a_{2} \ge 1$$ by AM-GM we get $$a_{1}+a_{2} \ge 2\sqrt{a_{1}a_{2}} \ge 2$$. (base case proven).

Now assume it is true for $$n=k$$. We will prove for $$n=k+1$$.

$$S_{k+1} = S_{k} + a_{k+1} \ge k + a_{k+1}$$

if $$a_{k+1} \ge 1$$ then the the proof is finished. Now if $$0 < a_{k+1} < 1$$, here is another approach in the induction: notice that the known inequality at the top of the post is equivalent with $$a_{k} \ge k/a_{k+1} - (k-1)/a_{k}$$ summing all from $$k=1,2,...,m$$ we get $$S_{m} \ge m/a_{m+1}$$ using this to prove for $$n=k+1$$ with $$0 < a_{k+1} < 1$$, we get

$$S_{k+1} = S_{k} + a_{k+1} \ge k/a_{k+1} + a_{k+1} = (k-1)/a_{k+1} + ( a_{k+1} + 1/a_{k+1} )$$

Now $$a_{k+1} + 1/a_{k+1} \ge 2$$ this is because $$f(x) = x + 1/x \ge 2, \:\: 0 < x < 1$$ (function is monotonically decreasing with convergence to 2). So we have

$$S_{k+1} \ge (k-1)/a_{k+1} + ( a_{k+1} + 1/a_{k+1} ) \ge (k-1) + 2 = k + 1$$

THus we have solved the problem.

• Unfinished approach:

Here is a hint of the IMO 2015 shortlisted problem:

"Using AM-GM on $$S_{k}$$ and $$k a_{k+1}$$ we may prove: $$S_{k} + k a_{k+1} \ge 2k$$ then sum all of them from $$k=1,2,....,m$$."

the inequality is quite easy to prove, but after the summation idk what else to do:

$$S_{1} + a_{2} \ge 2$$ $$S_{2} + 2 a_{3} \ge 2(2)$$ $$...$$ $$S_{m} + (m) a_{m+1} \ge 2(m)$$

then $$S_{1} + ... + S_{m} + a_{2} + ... + (m) a_{m+1} \ge m (m+1)$$

Remember $$S_k=a_1+a_2+\dots+a_k$$, so $$S_1+S_2+\dots+S_m=ma_1+(m-1)a_2+\dots+a_m.$$ Hence you have $$m(a_1+a_2+\dots+a_m+a_{m+1})\geq m(m+1)$$ which gives $$S_{m+1}\geq m+1$$.