Demostrar que para cualquier $n$,$\sum_{i=1}^n \frac{a_i}i \ge a_n $.

Question

Considerar la secuencia de $(a_n)_{n \ge1}$ real nos tal que $$a_{m+n}\le a_m+a_n,\ \forall m,n \ge 1$$

Demostrar que para cualquier $n$, $$\sum_{i=1}^n \frac{a_i}i \ge a_n .$$

Answer 1

Dada nuestra criterios $$a_u+a_v\geq a_{u+v},\ \forall u,v\geq1$$ we can substitute $u=k,\ v=n-k$ and get the equivalent $$a_k + a_{n-k} \geq a_{n}, \ \forall k \geq 1, \forall n \geq 2$$ which is a better starting point. Now we can continue with $$a_k + a_{n-k} \geq a_{n} \Leftrightarrow\\ \frac{a_k + a_{n-k}}{k} \geq \frac{a_{n}}{k} \Leftrightarrow\\ \sum_{k=1}^{n-1}(\frac{a_k + a_{n-k}}{k}) = \sum_{k=1}^{n-1}\frac{a_n}{k}$$ where we've simply divided by $k$ and then summed up both sides. Next we have $$\sum_{k=1}^{n-1}(\frac{a_k + a_{n-k}}{k}) = \sum_{k=1}^{n-1}\frac{a_k}{k} + \underbrace{\sum_{k=1}^{n-1}\frac{a_{n-k}}{k}}_{\text{ref 1.}} \geq \sum_{k=1}^{n-1}\frac{a_{n}}{k} = a_{n}\sum_{k=1}^{n-1}\frac{1}{k} \geq a_{n}\sum_{k=1}^{n-1}\frac{1}{n-1} = a_{n} \Leftrightarrow\\ \sum_{k=1}^{n-1}\frac{a_k}{k} + \underbrace{\sum_{k=1}^{n-1}\frac{a_{n-k}}{k}}_{\text{ref 1.}} \geq a_{n}$$ utilizing the know result that the harmonic series follows $\sum_{k=1}^n\frac{1}{k} \geq \sum_{k=1}^n\frac{1}{n} = 1$. To finish we have to show that $\text{ref 1.}$ is equal or greater than the missing term $\frac{a_n}{n}$ in the preceeding sum. $$\sum_{k=1}^{n-1}\frac{a_{n-k}}{k} = \frac{a_{n-1}}{1} + \frac{a_{n-2}}{2} + \cdots + \frac{a_{1}}{n-1} \geq\\ \frac{a_{n-1}}{n-1} + \frac{a_{n-2}}{n-1} + \cdots + \frac{a_{1}}{n-1} \geq \frac{a_n}{n-1}\cdot(\frac{n-1}{2})$$ once again using the result for the harmonic series and then our initial criteria in the last inequality, giving us $\frac{n-1}{2}$ identical terms. (If this fraction isn't an integer, thus the sum having an odd number of terms, the result is still the same. That's because the term in the middle of the sum is $a_{n/2}\geq \frac{a_n}{2}$ due to our base criteria yielding $a_{n/2}+a_{n/2}\geq a_n$, and the equation would've been $\frac{a_n}{n-1}\cdot(\frac{n-2}{2})+\frac{a_n}{(n-1)2}$.) Simplifying this gives $$\frac{a_n}{2} \geq \frac{a_n}{n}, \ \forall n \geq 2$$ which concludes the proof. $\Caja$