反向柯西应该可以

Maybe Reverse Cauchy can help us but I think we can use Rearrangement inequality in a very useful way here.

Guess the maximum value ...

与n的奇偶性有关

Exactly we will have two different cases with odd and even parities of $n$ and I send solution to sqing.

adityaguharoy wrote: Exactly we will have two different cases with odd and even parities of $n$ and I send solution to sqing. Thank adityaguharoy.

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.

It's an easy convexity.

arqady wrote: It's an easy convexity. Yes convexity. But not very easy. Okay yes it is easy .....

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

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 ??

Can you give your solution adityaguharoy??

See here : http://www.artofproblemsolving.com/community/c6h1064117p7435105

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

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.

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.

Can't we use titu lemma

Mate007 wrote: Can't we use titu lemma Can you show the solution with Titu's Lemma ?

It was also here: https://math.stackexchange.com/questions/2031113

Titus lemma would be helpful

Can you prove it with Titus Lemma ?

Special case of $n$ is even. https://artofproblemsolving.com/community/c6h1845556p12424479

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.

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?

Take double derivative with respect to each variable and show that it is $ \ge 0$.

Wrong Solution.

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.