trito11 wrote: a)Let$a,b,c\in\mathbb{R}$ and $a^2+b^2+c^2=1$.Prove that: $$|a-b|+|b-c|+|c-a| \le2\sqrt{2}$$b) Let $a_1,a_2,..a_{2019}\in\mathbb{R}$ and $\sum_{i=1}^{2019}a_i=1$.Find maximum of: $S=|a_1-a_2|+|a_2-a_3|+...+|a_{2019}-a_1|$ a) Assume $a\ge b\ge c.$ $$(|a-b|+|b-c|+|c-a|)^2=4(a-c)^2\leq 8( a^2+c^2)\leq 8.$$$$|a-b|+|b-c|+|c-a|\le2\sqrt{2}$$Let$a,b,c,d\geq 0$ and $a^2+b^2+c^2+d^2=1$.Prove that$$ |a-b|+|b-c|+|c-d|+|d-a| \le2$$

Not very difficult，we can use induction.

a) You just let a>/b>/c so the value of expression is 2(a-c) </ Sqrt2(a^2+c^2) but a^2 +B^2+ C^2</1 so 2(a-c) </2 sqrt 2 b) We will prove (A k - A k-1)(A k - A K+1) </0 with a2020=a0 and a2019=a1

I guess that nobody has posted complete solution. I claim that the answer is $2\sqrt{2018},$ that can be achieved when $a_i = \frac{(-1)^{i+1}}{\sqrt{2018}}$ for $1 \le i \le 2018$ and $a_{2019}=0$. To prove that it is the best constant, first notice that since $2019$ is odd, there is $1 \le i \le 2019$ such that $a_i$ and $a_{i+1}$ have the same sign (indices are considered mod $2019$), without loss of generality $a_{2019}$ and $a_1$. Denote by $S$ the whole expression. By $|x-y| \le |x| + |y|$ for all the terms except the last one, we get $S \le |a_{2019}-a_1| + |a_1| + |a_{2019}| + 2(|a_2| + |a_3| + ... + |a_{2018}|)$. It is clear that $|a_{2019}-a_1| + |a_1| + |a_{2019}| \le 2max(|a_1|, |a_{2019}|)$ since they have the same sign, without loss of generality $\le 2|a_1|$. Now $S\le 2(|a_1| + |a_2|+..+|a_{2018}|) \le 2 \sqrt{2018(a_1^2 + a_2^2 + ... + a_{2018}^2)} \le 2\sqrt{2018},$ and we are done. Fun fact: if we change $2019$ with $n,$ the answer is $2\sqrt{n-1}$ when $n$ is odd and $2\sqrt{n}$ for even $n$. Happy New Year to everyone!

trito11 wrote: a)Let$a,b,c\in\mathbb{R}$ and $a^2+b^2+c^2=1$.Prove that: $|a-b|+|b-c|+|c-a|\le2\sqrt{2}$ b) Let $a_1,a_2,..a_{2019}\in\mathbb{R}$ and $\sum_{i=1}^{2019}a_i^2=1$.Find the maximum of: $S=|a_1-a_2|+|a_2-a_3|+...+|a_{2019}-a_1|$

Attachments: