# 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?

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.

• What have been your conclusions and what is the source or the problem? May 17, 2021 at 18:14
• @PierreCarre No, because the statement would be false with $a_i=1$ for all $i$. May 17, 2021 at 18:25
• @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 May 17, 2021 at 18:28
• My initial though was Holder’s inequality, because of the different exponents, but I couldn’t get that to work. May 17, 2021 at 18:35
• There is an elementary proof, namely they provided an explicit counterexample to the inequality. What else are you asking for? May 20, 2021 at 0:19

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.

• 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. May 17, 2021 at 20:28
• @1qwertyyy I agree. Either there is an elementary approach or it is a very badly conceived question. But that can happen sometimes... May 17, 2021 at 20:31
• What do you mean by "badly conceived"? May 17, 2021 at 20:38
• @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. May 17, 2021 at 20:41
• @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 $f(a)>8$. May 17, 2021 at 20:53

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.

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.

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

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.

• 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 $1/4^2$. May 17, 2021 at 19:39
• @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! May 17, 2021 at 20:10
• what is the first inequality called? some generalization of rms-am? May 17, 2021 at 22:10
• @kyary: It's the Jensen inequality for the convex function $x\mapsto x^{3/2}$, slightly disguised May 17, 2021 at 22:17
• @kyary: It is also known as power mean inequality May 18, 2021 at 8:24