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As in the title

Is it true that $a_1^2a_2^2+a_2^2a_3^2+\cdots+a_8^2a_1^2 \leq 8$ if $a_1^3+a_2^3+\cdots+a_8^3=8 $ for positive numbers?

I'm not sure how to answer it. I've got some conclusion but those do not help, so any help would be appreciated.

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    $\begingroup$ What have been your conclusions and what is the source or the problem? $\endgroup$
    – VIVID
    May 17, 2021 at 18:14
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    $\begingroup$ @PierreCarre No, because the statement would be false with $a_i=1$ for all $i$. $\endgroup$
    – saulspatz
    May 17, 2021 at 18:25
  • $\begingroup$ @VIVID Once in question appears "for positive numbers" I instantly think of trying inequalities between means. Also, of course, every $a_i$ is less than two. Source: omj.edu.pl/uploads/attachments/om2017-tresci.pdf $\endgroup$
    – 1qwertyyyy
    May 17, 2021 at 18:28
  • $\begingroup$ My initial though was Holder’s inequality, because of the different exponents, but I couldn’t get that to work. $\endgroup$ May 17, 2021 at 18:35
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    $\begingroup$ There is an elementary proof, namely they provided an explicit counterexample to the inequality. What else are you asking for? $\endgroup$
    – Calvin Lin
    May 20, 2021 at 0:19

4 Answers 4

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The inequality does not hold. Try $$a = (0.150104,0.560857,1.0906,1.429,1.37785,0.967096,0.437358,0.0989485).$$

I obtained this vector as one of the candidates to min/max coming from the Lagrange multipliers method.

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  • $\begingroup$ Thank you for your answer. However I am sure there is more elementary approach for that statement since this question is supposed to be for middle/high schoolers. $\endgroup$
    – 1qwertyyyy
    May 17, 2021 at 20:28
  • $\begingroup$ @1qwertyyy I agree. Either there is an elementary approach or it is a very badly conceived question. But that can happen sometimes... $\endgroup$ May 17, 2021 at 20:31
  • $\begingroup$ What do you mean by "badly conceived"? $\endgroup$
    – 1qwertyyyy
    May 17, 2021 at 20:38
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    $\begingroup$ @1qwertyyyy If a problem is conceived for a certain set of students but cannot be solved with the tools available to them, it was not properly designed. $\endgroup$ May 17, 2021 at 20:41
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    $\begingroup$ @1qwertyyyy I see... I guess students could try to solve simpler problems where they fix some of the variables and play with the others. For instance, if you set $a_1=a_2=a_3=a_4=1.1$ and $a_5=a_6=a_7=0.8$, you can compute $a_8$ such that the restriction is satisfied and observe that $f(a)>8$. $\endgroup$ May 17, 2021 at 20:53
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The basic idea is that big jumps are bad - in particular jumping from above 1 to below 1.

The simplest way to achieve this is $(1,x,x,x,1,y,y,y)$ where $x^3+y^3=2$ and $x\neq 1$. We need to maximize $2x^4+2x^2 + 2y^4+2y^2$. By AMGM, when $x\neq 1$, $2x^4+2x^2> 4x^3$, so $2x^4+2x^2 + 2y^4+2y^2>4x^3+4y^3=8$ giving the desired elementary counterexample.

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Let $$f(x,y,z)=x^2y^2+y^2z^2+z^2x^2,\tag1$$ $$F(\overrightarrow a) = a_1^2a_2^2+a_2^2a_3^2+a_3^2a_4^2+\dots+a_8^2a_1^2,$$ then $$F(\overrightarrow a) = f(a_8,a_1,a_2) + f(a_2,a_3,a_4) + f(a_4,a_5,a_6) + f(a_6,a_7,a_8) - (a_2^2+a_6^2)(a_4^2+a_8^2).$$ If $$a_4=a_8\to 0,\quad a_2=a_6 = p,\quad a_1=a_3=a_5=a_7=q,\tag2$$ then $$F(\overrightarrow a) = 4p^2q^2\bigg|_{p^3+2q^3=4}=4\sqrt[3]4\,\left(q\sqrt[3]{2-q^3}\right)^2 = g(q),$$ where $$g' = 8\sqrt[3]4q\left(\sqrt[3]{2-q^3}\right)^2\left(1-\dfrac{q^3}{2-q^3}\right) , $$ $$\max F(\overrightarrow a) = 8\sqrt[3]2\quad\text{at}\quad q=1,\quad p=\sqrt[\large3]2.\tag3$$

Therefore, the OP bound should be increased.

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  • $\begingroup$ $4p^2q^2=4 *4^{1/3}$ when q is 1, p is $2^{1/3}$. I think there’s an error in your calc. $\endgroup$
    – Eric
    May 25, 2021 at 15:24
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A possible approach:

First, note that $$\left(\frac{\sum a_i^2 a_{i+1}^2} { 8}\right) ^{1/2} \le \left(\frac{\sum a_i^3 a_{i+1}^3} { 8}\right) ^{1/3}$$

Now, it is enough to check that $$\sum_{i=1}^8 a_i^3 a_{i+1}^3 \le 8$$ if $a_i$ are positive with $\sum a_i^3 =1$, or, equivalently $$\sum_{i=1}^8 b_i b_{i+1} \le 8$$

if $b_i$ are positive with sum $8$. Now, it is enough to check that

$$ \frac{\sum_{i=1}^ 8 x_i x_{i+1} }{ 8 } \le \left ( \frac{\sum x_i}{8} \right )^2 $$

for $x_i$ positive ( and say $\le 1$) .

$Added:$ It turns out that the last inequality is not correct, as @Martin R: pointed out. It may work for some smaller values of $n$, but not $n=8$.

Now it seems that the original inequality is also not correct.

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  • $\begingroup$ Unless I am mistaken, the last inequality does not hold for $x=(1, 1, 0, \ldots, 0)$: The LHS is $1/8$, and the RHS is $1/4^2$. $\endgroup$
    – Martin R
    May 17, 2021 at 19:39
  • $\begingroup$ @Martin R: Yes, you are righ! It's only true for a few small values of $n$, but not for $n=8$. Thanks for the input! $\endgroup$
    – orangeskid
    May 17, 2021 at 20:10
  • $\begingroup$ what is the first inequality called? some generalization of rms-am? $\endgroup$
    – kyary
    May 17, 2021 at 22:10
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    $\begingroup$ @kyary: It's the Jensen inequality for the convex function $x\mapsto x^{3/2}$, slightly disguised $\endgroup$
    – orangeskid
    May 17, 2021 at 22:17
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    $\begingroup$ @kyary: It is also known as power mean inequality $\endgroup$
    – Martin R
    May 18, 2021 at 8:24

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