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give the postive intger $n\ge 2$,and postive real numbers $a<b$ if the real numbers such $x_{1},x_{2},\cdots,x_{n}\in[a,b]$ find the maximum of the value $$\dfrac{\frac{x^2_{1}}{x_{2}}+\frac{x^2_{2}}{x_{3}}+\cdots+\frac{x^2_{n-1}}{x_{n}}+\frac{x^2_{n}}{x_{1}}}{x_{1}+x_{2}+\cdots+x_{n}}$$

it seem the polya-szego inequality http://journalofinequalitiesandapplications.springeropen.com/articles/10.1186/1029-242X-2013-591

Michael Rozenberg
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math110
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2 Answers2

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Let $M$ is a maximum value (it exists because continuous function on compact gets there a maximum value) and $$f(x_1,x_2,...x_n)=\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+...+\frac{x_n^2}{x_1}-M(x_1+x_2+...+x_n).$$ Since $f$ is a convex function for all $x_i$, we obtain $$0=\max_{\{x_1,x_2,...,x_n\}\subset[a,b]}f=\max_{\{x_1,x_2,...,x_n\}\subset\{a,b\}}f$$ From here if $n$ is even we have $$M=\frac{\frac{a^2}{b}+\frac{b^2}{a}}{a+b}=\frac{a}{b}+\frac{b}{a}-1,$$ which occurs for $x_1=x_3=...=a$ and $x_2=x_4=...=b$.

If $n$ is odd we have for $n=2m+1$: $$M=\frac{m\left(\frac{a^2}{b}+\frac{b^2}{a}\right)+a}{(m+1)a+mb},$$ which occurs for $x_1=x_3=...=x_{2m+1}=a$ and $x_2=x_4=...=x_{2m}=b$.

Hans
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Michael Rozenberg
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  • I`d need some explanation about the convexity of the function, but I really don't understand the downvotes. – Rafa Budría Feb 21 '17 at 10:46
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    @Rafa Budría because $\frac{\partial^2x_i}{\partial x_i^2}=\frac{2}{x_{i+1}}+\frac{2x_{i-1}}{x_i^3}>0$ – Michael Rozenberg Feb 21 '17 at 12:37
  • Thank you! Still remains the other puzzling question ;-) – Rafa Budría Feb 21 '17 at 14:52
  • Could you please explain how you obtained the maximum for the discrete corner case, i.e., when $x_i\in{a,b}$? – Hans Feb 25 '17 at 02:02
  • @Hans For all $i$ we see that $f$ is a convex function of $x_i$. Thus, $f$ gets a maximum value for an extremal value of $x_i$, id est, for $x_i\in{a,b}$. . – Michael Rozenberg Feb 25 '17 at 04:39
  • You misunderstood me. I am asking about the maximum for the discrete corner case where $x_i\in{a,b}$, NOT the continuous case before its reduction to the discrete case. In other words, I am asking about how you reach the final conclusion how the $a$ and $b$ should be assigned to $x_i$. – Hans Feb 25 '17 at 04:46
  • @G Cab With your example, if $n=4$ we have $\frac{a^2-ab+b^2}{ab}\geq\frac{2a+\frac{a^2}{b}+\frac{b^2}{a}}{3a+b}$ is true because it's $a(a-b)^2\geq0$. – Michael Rozenberg Feb 25 '17 at 04:48
  • @Hans It's a very ugly work. – Michael Rozenberg Feb 25 '17 at 04:55
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    That is even more reason to show it. Otherwise it is pretty mysterious. Please do write it out in your answer. I would really appreciate it. – Hans Feb 25 '17 at 07:19
  • @MichaelRozenberg, you are right, sorry: I deleted my comment – G Cab Feb 26 '17 at 19:56
  • @MichaelRozenberg - In general, positivity of second derivatives does not guarantee convexity in higher dimensions (e.g., $f(x,y)=xy$) so your reply to Rafa Budria's comment is not complete. – uniquesolution Feb 27 '17 at 15:21
  • @uniquesolution I did not say that $f$ is a convex function. I said that $f$ is a convex function of $x_i$ for all $i$. – Michael Rozenberg Feb 27 '17 at 20:31
  • @uniquesolution: What you said was right. Michael Rozenberg is right that we only need convexity in each variable to reduce the problem to corner case. However, it happens that $f$ is indeed convex, since $\frac{x_i^2}{x_j}$ is convex $\forall i, j$ and the sum of them is convex while $-M\sum_i x_i$ is convex. – Hans Feb 28 '17 at 02:34
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    @MichaelRozenberg: Would you mind writing out your proof for the final discrete case that the continuous case has reduced to? I am extremely curious about it. Thank you! – Hans Mar 04 '17 at 20:42
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This is a proof of the discrete case for Michael Rozenberg's reply, assuming $n\ge 2$ and therefore $M>1$.

Consider a cyclic word $x$ on $\{a,b\}^n$ maximizing $f$. We can write $f(x)$ as a linear function of the counts of the $n$ cyclic bigrams $x_i x_{(i+1)\bmod n}$: $$f(x) = n_{aa} w(a,a) + n_{ab} w(a,b) + n_{ba} w(b,a) + n_{bb} w(b,b)$$ where $w(s,t)=\frac{s^2}{t} - M s$.

Because $M>1$, $w(b,b) < w(a,a)$ therefore $x\ne b^n$ and if $x$ contained an instance of $bb$, we could increase $f(x)$ by replacing it with $b$ and inserting another $a$ next to any $a$. Thus $n_{bb}=0$.

And because there are as many $ba$ bigrams as $ab$, we have: $$f(x) = \left(n-2n_{ab}\right) w(a,a) + n_{ab} \left(w(a,b)+w(b,a)\right)$$ This is a linear function of $n_{ab}$ so the maximum is reached when $x$ is either of the form $a^n$ or $(ab)^* a?$. The former is never maximal because $M>1$, therefore Michael Rozenberg's construction is optimal.

ffee
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