Let the numbers a_{0}, a_{1}, ... , a_{n} be given, where a_{0} = 1 and
the other numbers are arbitrary reals (n \geq 1). We prove the validity of
the following inequality:
\sum^{n-1}_{k=0} a_{k+1}(a_{k} - a_{k+1}) \leq \frac{n}{2(n+1)}.
Keeping in mind that a_{0} = 1, we obtain sequentially that:
\align \sum^{n-1}_{k=0}& a_{k+1}(a_{k} - a_{k+1}) = - \sum^{n}_{k=1}
a_{k}^{2} + \sum^{n-1}_{k=0} a_{n-k}a_{n-k-1} = -a^2_n - \sum^{n-1}_{k=1} a^2_{n-k} + \sum^{n-1}_{k=0} a_{n-k} a_{n-k-1}\\
&= - a^2_n - \sum^{n-1}_{k=1} \frac{(k+2)+k}{2(k+1)} a^2_{n-k} + \sum^{n-1}_{k=0} a_{n-k} a_{n-k-1}\\
&= \frac{n}{2(n + 1)}
- \sum^{n-1}_{k=0} \frac{k + 2}{2(k + 1)}a_{n-k}^{2}
- \sum^{n-1}_{k=0} \frac{k + 1}{2(k + 2)}a_{n-k-1}^{2}
+ \sum^{n-1}_{k=0} a_{n-k}a_{n-k-1}\\
&= \frac{n}{2(n + 1)}
- \sum^{n-1}_{k=0} (\frac{k + 2}{2(k + 1)}a_{n-k}^{2}
- a_{n-k}a_{n-k-1} + \frac{k + 1}{2(k + 2)}a_{n-k-1}^{2})\\
&= \frac{n}{2(n + 1)}
- \sum^{n-1}_{k=0} \frac{k + 2}{2(k + 1)}(a_{n-k}^{2}
- \frac{2(k + 1)}{k + 2}a_{n-k}a_{n-k-1}
+ (\frac{k + 1}{k + 2})^{2}a_{n-k-1}^{2})\\
&= \frac{n}{2(n+1)}
- \sum^{n-1}_{k=0} \frac{k + 2}{2(k+1)}(a_{n-k}
- \frac{k + 1}{k + 2}a_{n-k-1})^{2}
\leq \frac{n}{2(n + 1)} align
which proves the inequality.
Obviously, this inequality is a generalization of the well-known inequality
a(1 - a) \leq \frac{1}{4}.
Bulgarian Academy of Sciences
Honorary Associate, School of Mathematical Sciences
University of Technology, Sydney