Given an integer $n \geq2$ and real numbers $a,b$ such that $0<a<b$. Let $x_1,x_2,\ldots, x_n\in [a,b]$ be real numbers. Find the maximum value of $$\frac{\frac{x^2_1}{x_2}+\frac{x^2_2}{x_3}+\cdots+\frac{x^2_{n-1}}{x_n}+\frac{x^2_n}{x_1}}{x_1+x_2+\cdots +x_{n-1}+x_n}.$$
Problem
Source: China Mathematical Olympiad 2017, Problem 6
Tags: BPSQ, China, PRC, inequalities, Inequality proposed, convex-concave inequalities, sqing orz
24.11.2016 10:27
反向柯西应该可以
24.11.2016 10:33
Maybe Reverse Cauchy can help us but I think we can use Rearrangement inequality in a very useful way here.
24.11.2016 12:39
Guess the maximum value ...
24.11.2016 13:43
与n的奇偶性有关
24.11.2016 15:11
Exactly we will have two different cases with odd and even parities of $n$ and I send solution to sqing.
24.11.2016 15:47
adityaguharoy wrote: Exactly we will have two different cases with odd and even parities of $n$ and I send solution to sqing. Thank adityaguharoy.
25.11.2016 08:18
When $n$ is odd ,$\frac{\frac{x^2_1}{x_2}+\frac{x^2_2}{x_3}+\cdots+\frac{x^2_{n-1}}{x_n}+\frac{x^2_n}{x_1}}{x_1+x_2+\cdots +x_{n-1}+x_n}\leq \frac{\frac{n-1}{2}\cdot (\frac{a^2}{b}+\frac{b^2}{a})+a}{(\frac{n-1}{2}+1 ) \cdot a+\frac{n-1}{2}\cdot b }$ , When $n$ is even ,$\frac{\frac{x^2_1}{x_2}+\frac{x^2_2}{x_3}+\cdots+\frac{x^2_{n-1}}{x_n}+\frac{x^2_n}{x_1}}{x_1+x_2+\cdots +x_{n-1}+x_n}\leq \frac{\frac{a^2}{b}+\frac{b^2}{a}}{a+b}$. Thank adityaguharoy.
26.11.2016 11:26
It's an easy convexity.
28.11.2016 06:41
arqady wrote: It's an easy convexity. Yes convexity. But not very easy. Okay yes it is easy .....
02.12.2016 18:46
From $\frac{x_1^2}{x_2}\le x_1(\frac{a}{b}+\frac{b}{a})-x_2$ So, $\frac{x_1^2}{x_2}+...+\frac{x_n^2}{x_1}\le \frac{a^2-ab+b^2}{ab}(x_1+...+x_n)$ Then, maximum is $\frac{a^2-ab+b^2}{ab}$ Check me pls
02.12.2016 19:37
Gems98 wrote: From $\frac{x_1^2}{x_2}\le x_1(\frac{a}{b}+\frac{b}{a})-x_2$ So, $\frac{x_1^2}{x_2}+...+\frac{x_n^2}{x_1}\le \frac{a^2-ab+b^2}{ab}(x_1+...+x_n)$ Then, maximum is $\frac{a^2-ab+b^2}{ab}$ Check me pls How do you get that ? OK the answer is true for $n$ even but how do you get that ? From where do you get that ??
17.12.2016 07:32
Can you give your solution adityaguharoy??
13.01.2017 15:35
See here : http://www.artofproblemsolving.com/community/c6h1064117p7435105
16.03.2017 12:59
adityaguharoy wrote: See here : http://www.artofproblemsolving.com/community/c6h1064117p7435105 please can u show me why x_i are taking a and b alternatively to be maximum
17.03.2017 19:32
Erkhes wrote: adityaguharoy wrote: See here : http://www.artofproblemsolving.com/community/c6h1064117p7435105 please can u show me why x_i are taking a and b alternatively to be maximum Rearrangement.
18.03.2017 13:04
adityaguharoy wrote: Erkhes wrote: adityaguharoy wrote: See here : http://www.artofproblemsolving.com/community/c6h1064117p7435105 please can u show me why x_i are taking a and b alternatively to be maximum Rearrangement. i meant i really want to know how u using rearrangement.
19.12.2017 08:14
Can't we use titu lemma
19.12.2017 08:18
Mate007 wrote: Can't we use titu lemma Can you show the solution with Titu's Lemma ?
19.12.2017 09:17
It was also here: https://math.stackexchange.com/questions/2031113
19.12.2017 10:08
Titus lemma would be helpful
19.12.2017 19:25
Can you prove it with Titus Lemma ?
26.05.2019 14:13
Special case of $n$ is even. https://artofproblemsolving.com/community/c6h1845556p12424479
26.05.2019 14:17
arqady wrote: It was also here: https://math.stackexchange.com/questions/2031113 Yes the solution is nicely written there, by you. Congratulations and thank you @arqady sir.
27.05.2019 14:13
adityaguharoy wrote: arqady wrote: It's an easy convexity. Yes convexity. But not very easy. Okay yes it is easy ..... can someone explain why this function is convex?
28.05.2019 06:42
Take double derivative with respect to each variable and show that it is $ \ge 0$.
04.09.2021 07:01
Wrong Solution.
21.03.2022 03:44
Easy for #6 In this solution, all indices are taken mod $n$ and the key ingredient is rearrangement. When $n$ is even, the answer is $\frac ab + \frac ba - 1$, obtained when $x_{2i-1}=a$ and $x_{2i}=b$ for $i=1,\cdots,\frac n2$ For $n$ even we prove this is optimal with rearrangement inequality. We permute $x_1,\cdots,x_n$. This clearly doesn't affect the denominator. Let $z_j=x_j^2$ and $w_j=\frac{1}{x_j}$, and WLOG $z_1\le z_2\le \cdots \le z_n$ which implies $w_1\ge w_2\ge \cdots \ge w_n$. By rearrangement inequality, $\sum\limits_{j=1}^n w_jz_{j-1} \le \sum\limits_{j=1}^n w_{n+1-j}z_j = \sum\limits_{j=1}^{\frac n2} (\frac{x_j^2}{x_{n+1-j}} + \frac{x_{n+1-j}^2}{x_j})$. Observe $\frac{\frac{x_j^2}{x_{n+1-j}} + \frac{x_{n+1-j}^2}{x_j}}{x_{n+1-j}+x_j} = \frac{x_j}{x_{n+1-j}} + \frac{x_{n+1-j}}{x_j} - 1 \le \frac ab + \frac ba -1$ because $\frac ba \ge \frac{x_j}{x_{n+1-j}} \ge \frac ab$ This means $\sum\limits_{j=1}^{\frac n2} (\frac{x_j^2}{x_{n+1-j}} + \frac{x_{n+1-j}^2}{x_j})$ $\le \sum\limits_{j=1}^{\frac n2} (x_j+x_{n+1-j})(\frac ab + \frac ba - 1)=(\frac ab + \frac ba - 1)(\sum\limits_{j=1}^n x_j)$, proving the optimality of $\frac ab + \frac ba - 1$ It remains to settle $n$ odd. Let $n=2k+1$. I claim the answer is $\frac{a^2b+k(a^3+b^3)}{ab((k+1)a+kb)}$, obtained when $x_{2j-1}=a$ where $j=1,\cdots,k+1$ and $x_{2j}=b$ when $j=1,\cdots,k$ Let $C=\frac{a^2b+k(a^3+b^3)}{ab((k+1)a+kb)}$. Let $y_1\le y_2\le \cdots \le y_n$ such that the multisets $y_j$, $x_j$ are equal. By rearrangement inequality, it suffices to show $y_{k+1}(1-C)+\sum\limits_{j=1}^k (y_j+y_{2k+2-j})(\frac{y_j}{y_{2k+2-j}} + \frac{y_{2k+2-j}}{y_j} -1-C) \le 0$ Claim: When this expression is maximized, $y_{k+1}=a$ and $y_j=b$ for all $j=k+2,\cdots,2k+1$ Proof: Note $1-C<0$ so we want to minimize $y_{k+1}$. We can also assume $\frac{y_j}{y_{2k+2-j}} + \frac{y_{2k+2-j}}{y_j} -1-C>0$, or we can remove and induct down on $k-1$. If $y_j<b$ where $j\ge k+2$ we can replace $(y_{n+1-j},y_j)$ with $(b\frac{y_{n+1-j}}{y_j}, b)$. These numbers are still in range. Now recall $y_{k+1}\ge y_1,\cdots,y_k$ so we are done.