Does anyone have solution...?

Could someone post a solution to this challenging problem please (e.g. official)? Thank you very much!

Anyone have a solution?

The answer is probably $ \sqrt{p-1}\left [ \frac{p}{3} \right ] $.

How to derive the result of toto1234567890. Btw, I don't think it holds over all $p-1$ tuples

I don't understand this problem

Hint: Think of vectors.

How would one use vectors for this problem? Isn't vectors used for geometry? Thanks!

Could someone post a solution? Thanks!

Too difficult for me.

Bump. Any solution?

What it mean d(a,b)

$d(a,b)$ is defined in the first sentence: Quote: Given a prime $p$, let $d(a,b)$ be the number of integers $c$ such that $1 \leq c < p$, and the remainders when $ac$ and $bc$ are divided by $p$ are both at most $\frac{p}{3}$.

toto1234567890 wrote: Hint: Think of vectors. How could one use vector on this problem. Can somebody elaborate this idea?

Zhero wrote: Given a prime $p$, let $d(a,b)$ be the number of integers $c$ such that $1 \leq c < p$, and the remainders when $ac$ and $bc$ are divided by $p$ are both at most $\frac{p}{3}$. Determine the maximum value of \[\sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)(x_a + 1)(x_b + 1)} - \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)x_ax_b}\]over all $(p-1)$-tuples $(x_1,x_2,\ldots,x_{p-1})$ of real numbers. Brian Hamrick. Okay. so, lots of people ask for me to show the solution, so just an outline How to use vectors... First, let's say vector $ v_a $ in $ \mathbb{R}^{p-1} $ is has $ 0 $ in term $ c $ if remainder of $ ac $ is bigger than $ \frac{p}{3} $ and something(this should be $ x_a+1 $ or $ x_a $) if remainder of $ ac $ is at most $ \frac{p}{3} $ . Then we could see that $ \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)(x_a + 1)(x_b + 1)} $ and $ \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)x_ax_b} $ are some size of a vector and now, triangle inequality kills this and gets $ \sqrt{p-1}\left [ \frac{p}{3} \right ] $ the answer.

Let $f(a,b)=\begin{cases}1&ab\mod p<p/3\\0&\text{otherwise}\end{cases}.$ Then $$d(a,b)=\sum_{c=1}^{p-1}f(a,c)f(b,c).$$So by Minkowski Inequality, \begin{align*}\sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)(x_a + 1)(x_b + 1)} - \sqrt{\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}d(a,b)x_ax_b}&=\sqrt{\sum_{c=1}^{p-1}\left(\sum_{a=1}^{p-1}(x_a + 1)f(a,c)\right)^2} - \sqrt{\sum_{c=1}^{p-1}\left(\sum_{a=1}^{p-1}x_a f(a,c)\right)^2}\\&\le\sqrt{\sum_{c=1}^{p-1}\left(\sum_{a=1}^{p-1}f(a,c)\right)^2}=\sqrt{p-1}\left \lfloor \frac{p}{3} \right \rfloor.\Box\end{align*}Equality holds when all $x_i=0.\Box$