Find the maximum of $\prod_{cyc}(a^2_{1}+a^2_{2})$

Let $$n$$ be an odd number, and $$a_{i}\ge 0$$ such that $$a_{1}+a_{2}+\cdots+a_{n}=n$$ Find the maximum of the value $$f(n)=(a^2_{1}+a^2_{2})(a^2_{2}+a^2_{3})\cdots(a^2_{n}+a^2_{1})$$

Now I have solved the case $$n=3$$: Let $$a_{3}=\min(a_{1},a_{2},a_{3})$$, then we have $$a^2_{2}+a^2_{3}\le \left(a_{2}+\dfrac{a_{3}}{2}\right)^2=x^2$$ $$a^2_{1}+a^2_{3}\le \left(a_{1}+\dfrac{a_{3}}{2}\right)^2=y^2$$ $$a^2_{1}+a^2_{2}\le x^2+y^2$$ where $$x=a_{2}+\dfrac{a_{3}}{2},y=a_{1}+\dfrac{a_{3}}{2},x+y=3$$, so $$f(3)\le x^2y^2(x^2+y^2)=\dfrac{1}{2}xy\cdot 2xy(x^2+y^2)\le \dfrac{1}{2}\dfrac{(x+y)^2}{4}\dfrac{(x+y)^4}{4}=\dfrac{3^6}{32}$$ with equality when $$x=y=\dfrac{3}{2}$$ or $$a_{1}=a_{2}=\dfrac{3}{2},a_{3}=0$$.

• Yes, for $n=4$ it's easy. – Michael Rozenberg Feb 11 at 9:17
• Ok, i forgot this term. – Dr. Sonnhard Graubner Feb 11 at 9:17
• $n$ should be odd Michael. – Dr. Sonnhard Graubner Feb 11 at 9:18
• So $$(a_1^2+a_2^2)(a_2^2+a_3^2)(a_3^2+a_1^2)\le \frac{729}{32}$$ – Dr. Sonnhard Graubner Feb 11 at 9:20
• @Sonnhard Have you a solution for $n=4$? – Michael Rozenberg Feb 11 at 10:13

Let us fix an integer $$n \ge 2$$ (the following results are not restricted to odd numbers). The function $$F(x_1, \ldots, x_n) = (x_1^2 + x_2^2)(x_2^2 + x_3^2) \cdots (x_n^2 + x_1^2)$$ is continuous on the compact set $$K = \{ x \in \Bbb R_{\ge 0}^n \mid \sum_{k=1}^n x_k = n \}$$ and therefore attains its maximum at some point $$a =(a_1,\ldots, a_n) \in \Bbb R_{\ge 0}^n$$: $$\tag{M} f(n) = \max \{ F(x) \mid x \in K \} = F(a) \, .$$ The idea is to show that

\begin{aligned} a &= (0, 2, 0, 2, \ldots, 0, 2) \quad\text{if n is even,} \\ a &= (0, 2, 0, 2, \ldots, 0, 3/2, 3/2) \quad \text{if n is odd.} \end{aligned} \tag{*} or a cyclic permutation thereof.

It then follows that $$f(n) = \begin{cases} 2^{2n} & \quad\text{if n is even,} \\ 2^{2n-11} \cdot 3^6 & \quad\text{if n is odd.} \end{cases}$$

The proof of $$(*)$$ is done in several steps. In each step some property of the “maximal vector” $$a$$ is obtained. This is always done by showing that if $$a$$ does not have this property then there is another vector $$\tilde a$$ with the same sum of its components, and $$F(\tilde a) > F(a)$$, in contradiction to the maximality condition $$(M)$$.

Step 1: $$a_k > 0 \quad \implies \quad a_{k-1} < a_k \text{ or } a_{k+1} < a_k \,.$$ In particular, $$\min(a_1, \ldots, a_n) = 0$$.

Roughly speaking: a nonzero component cannot be a “local minimum.”

Proof: Assume that $$\min(a_{k-1}, a_{k+1}) \ge a_k > 0$$. (Here and in the following, all index calculations are done $$\bmod n$$.) Then $$(a_{k-1}^2 + a_k^2)(a_k^2 + a_{k+1}^2) \le (a_{k-1}^2 + a_{k-1} a_k)(a_k a_{k+1} + a_{k+1}^2) \\ < (a_{k-1} + \frac 12 a_k)^2 (a_{k+1} + \frac 12 a_k)^2$$ so that $$F(\ldots, a_{k-1}, a_k, a_{k+1},\ldots) < F(\ldots, a_{k-1} + \frac 12 a_k, 0, a_{k+1} + \frac 12 a_k, \ldots)$$ in contradiction to the maximality condition $$(M)$$.

Step 2: \begin{aligned} &0 = a_{k-1} < a_k \le a_{k+1} \le a_{k+2} \quad \text{or} \\ &a_k \ge a_{k+1} \ge a_{k+2} > a_{k+3} = 0 \end{aligned} does not hold for any index $$k$$.

In other words: a zero component cannot be followed by three increasing nonzero components, or preceded by three decreasing nonzero components.

Proof: Assume that $$0 = a_{k-1} < a_k \le a_{k+1} \le a_{k+2}$$. Then $$a_k^2 ( a_k^2+ a_{k+1}^2)(a_{k+1}^2 + a_{k+2}^2) = a_k^4 a_{k+1}^2 + a_k^4 a_{k+2}^2 + a_k^2 a_{k+1}^4 + a_k^2 a_{k+1}^2 a_{k+2}^2 \\ \le a_k^4 a_{k+2}^2 + 3 a_k^2 a_{k+1}^2 a_{k+2}^2 < (a_k + a_{k+1})^4 a_{k+2}^2$$ so that $$F(\ldots, 0, a_k, a_{k+1},a_{k+2}, \ldots) < F(\ldots, 0, a_k + a_{k+1}, 0, a_{k+2}, \ldots)$$ in contradiction to the maximality condition $$(M)$$.

Step 3: At most two consecutive components $$a_k$$ are nonzero.

Proof: Let $$a_k, \ldots, a_{k+l-1}$$ be a maximal range of $$l$$ consecutive nonzero components of $$a$$: \begin{aligned} a_{k-1} &= 0 \\ a_j &> 0 \quad \text{for j=k, \ldots, k+l-1}, \\ a_{k+l} &= 0 \end{aligned} Let $$m \in \{ k, \ldots, k+l-1 \}$$ be an index where the maximum of those components is attained: $$a_m = \max \{ a_k, \ldots, a_{k+l-1} \} \, .$$ From Step 1 it follows that the components are (weakly) increasing up to $$a_m$$, and (weakly) decreasing after $$a_m$$: $$0 = a_{k-1} < a_k \le a_{k+1} \le \ldots \le a_{m-1}\le a_m \\ \ge a_{m+1} \ge \ldots \ge a_{k+l-1} > a_{k+l} = 0 \, .$$ From Step 2 it then follows that $$a_m$$ is preceded and followed by at most one nonzero component, which means that $$l \le 3$$.

It remains to exlude the case $$l=3$$: Assume that $$0 = a_{k-1} < a_k \le a_{k+1} \ge a_{k+2} >a_{k+3} = 0 \, .$$ Then $$a_k^2 ( a_k^2 + a_{k+1}^2) < (a_k + \frac 12 a_{k+1})^4 \, ,\\ (a_{k+1}^2 + a_{k+2}^2) a_{k+1}^2 < (a_{k+2} + \frac 12 a_{k+1})^4 \, ,$$ so that $$F(\ldots, 0, a_k ,a_{k+1},a_{k+2},0,\ldots) < F(\ldots,0, a_k + \frac 12 a_{k+1},0, a_{k+2} + \frac 12 a_{k+1}, 0, \ldots) \, .$$

So up to now we have shown that the components of $$a$$ consist of “blocks,” starting with zero and followed by one or two nonzero components. These blocks can be rearranged without changing the value $$F(a)$$. Therefore we can assume that $$a$$ consists of zero or more “2-blocks” followed by zero or more “3-blocks:” $$a = (0, u_1, \ldots, 0, u_N, 0, v_1,w_1, \ldots, 0, v_M, w_M)$$

Step 4: $$v_j = w_j$$ in each 3-block.

Proof: If $$v_j \ne w_j$$ then $$v_j^2(v_j^2 + w_j^2)w_j^2 < 2 \left( \frac{v_j+w_j}2 \right)^6$$ so that $$F(a)$$ increases if the block $$(0,v_j,w_j)$$ is replaced by $$(0,\frac{v_j+w_j}2,\frac{v_j+w_j}2)$$.

Step 5: All 3 blocks are identical: $$v_1 = \ldots = v_M$$.

Proof: Each 3 block contributes $$2v_j^6$$ to the product $$F(a)$$. The total contribution of all 3-blocks is $$2^M (v_1 \cdots v_M)^6 \le 2^M \left( \frac{v_1 + \ldots + v_M}{M}\right)^{6M}$$ with equality if and only if all $$v_j$$ are equal (AM-GM inequality).

Step 6: There is at most one 3-block.

Proof: Replacing two 3-blocks $$(0,v, v, 0,v, v)$$ by three 2-blocks $$(0,u,0,u,0, u)$$ with $$u = \frac 43 v$$ preserves the sum of the components, but increases the value of $$F(a)$$.

Step 7: All 2-blocks are identical: $$u_1 = \ldots = u_N$$.

The proof is identical to that of Step 5.

Finale: For even $$n$$ we are done. $$a$$ consists only of (identical) 2-blocks, and the sum of the components is $$n$$, i.e. $$a = (0, 2, 0, 2, \ldots , 0,2) \,.$$ If $$n$$ is odd then $$(n-3)/2$$ 2-blocks are followed by a single 3-block $$a = (0,u,0,u,\ldots,0, v, v) \quad \text{where }\frac{n-3}{2}u + 2v = n \, .$$ Then $$F(a) = 2 u^{2n-6}v^6 = 2 \left(\frac 43 \right)^{2n-6} \left(\frac 34 u\right)^{2n-6}v^6 \, .$$ The AM-GM inequality shows that $$\left(\frac 34 u\right)^{2n-6}v^6 \le \left( \frac{(2n-6)\frac 34 u + 6 v}{2n}\right)^{2n} = \left( \frac{3n}{2n}\right)^{2n} = \left( \frac{3}{2}\right)^{2n} \, .$$ with equality if and only if $$\frac 34 u = v$$. It follows that $$u=2$$ and $$v= \frac 32$$ for the maximal vector $$a$$, i.e. $$a = (0,2,0,2,\ldots,0, 3/2, 3/2) \, ,$$ and that finishes the proof of $$(*)$$.