How Can We Prove the Summation Inequality for a Given Sequence?

[anemone]

Assume that $x_1,\,x_2,\,\cdots,\,x_n \ge -1$ and $\displaystyle \sum_{i=1}^n x_i^3=0$. Prove that $\displaystyle \sum_{i=1}^n x_i \le \dfrac{n}{3}$.

 

[jvicens]

Thank you for sharing your problem with us. I would like to provide a proof for the statement you have proposed.

We are given that $x_1,\,x_2,\,\cdots,\,x_n \ge -1$ and $\displaystyle \sum_{i=1}^n x_i^3=0$. Our goal is to prove that $\displaystyle \sum_{i=1}^n x_i \le \dfrac{n}{3}$.

First, let us consider the case when $n=1$. In this case, we have $x_1^3=0$, which implies that $x_1=0$. Therefore, $\displaystyle \sum_{i=1}^n x_i = x_1 = 0 \le \dfrac{1}{3} = \dfrac{n}{3}$.

Now, let us assume that the statement holds for $n=k$ and prove it for $n=k+1$. This means that for $n=k$, we have $\displaystyle \sum_{i=1}^k x_i \le \dfrac{k}{3}$. Adding $x_{k+1}$ to both sides, we get $\displaystyle \sum_{i=1}^{k+1} x_i \le \dfrac{k}{3} + x_{k+1}$. Since $x_{k+1} \ge -1$, we can write $\dfrac{k}{3} + x_{k+1} \le \dfrac{k}{3} + \dfrac{1}{3} = \dfrac{k+1}{3} = \dfrac{n}{3}$. Therefore, for $n=k+1$, we also have $\displaystyle \sum_{i=1}^{k+1} x_i \le \dfrac{n}{3}$.

By the principle of mathematical induction, we can conclude that for any positive integer $n$, $\displaystyle \sum_{i=1}^n x_i \le \dfrac{n}{3}$.

I hope this proof helps to clarify the statement and its validity. If you have any further questions or concerns, please do not hesitate to ask.