Product of two vector inequality Given $0<a_1\leq \dots a_n\leq 1$ and $0\leq b_1 \leq \dots b_n \leq 1$, s.j. $\sum_{i=1}^n a_i=1$
Can we prove: $\sum_{i=1}^n a_ib_i \geq \sum_{i=1}^n \frac{1}{n} b_i$ ?
 A: That comes from the Chebyshev sum inequality:

If $a_1\le\dots\le a_n$ and $b_1\le \dots\le b_n$, then
  $$n\sum_{i=1}^na_ib_i\ge\sum_{i=1}^na_i\cdot\sum_{i=1}^nb_i$$

which is itself a consequence of the rearrangement inequality:

Under the same hypotheses, the sum
  $$\sum_{i=1}^na_ib_{\sigma(i)},\quad \sigma \in S_n$$
  is maximal when $\sigma=\operatorname{id}$.

A: $\sum_i a_ib_i$ is equivalent to $a^\top b$, which is equivalent to $\|a\|\cdot\|b\|\cos\angle(a,b)$.
$\sum_i b_i/n$ is equivalent to $v^\top b$, where $v$ is a constant $1/n$ vector.
Then, you know that $\|v\|\le \|a\|$ and $\cos\angle(v,b)\le\cos\angle(a,b)$.
For the first part, you have:
$$
\|v\|^2=\sum_i \frac{1}{n^2}=\frac{1}{n}
$$
Then, $\|a\|^2=\sum_i a_i^2$. But the simplest change from $v$ to $a$ consist in adding $e$ to a component and remove $e$ to another component. Any further change can be done the same way.
So we get:
\begin{align}
\|a\|^2&=\frac{n-2}{n^2}+\left(\frac{1}{n}+e\right)^2+\left(\frac{1}{n}-e\right)^2\\
&=\frac{n-2}{n^2}+2\frac{1}{n^2}+2e^2+2\frac{e}{n}-2\frac{e}{n}\\
&=\frac{1}{n}+\gamma,
\end{align}
where $\gamma\ge 0$
For the second part, basically, the components of the vectors $a$ and $b$ are non-decreasing. While the components of $v$ are constant. So $a$ and $b$ are more in the same direction. Hence, the cosine is greater.
In the end, you have $a^\top b\ge v^\top b$.
