Show that $\sum^{6}_{i=1} a_{i}=\frac{15}{2}$ and $ \sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4} \implies \prod_{i=1}^{6} a_{i} \leq \frac{5}{2}$

Question

Let $a_{i}$, $1 \leq i \leq 6,$ be real numbers such that

$\displaystyle\hspace{1.2 in}\sum^{6}_{i=1} a_{i}=\frac{15}{2}\;\;$ and $\;\;\displaystyle\sum^{6}_{i=1} a^{2}_{i}=\frac{45}{4}$.

Prove that $\hspace{.15 in}\displaystyle\prod_{i=1}^{6} a_{i} \leq \frac{5}{2} $.

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I was thinking if I consider the first summation and extended it, it going be pretty long which $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}= \frac{15}{2}$ and the second one like $a^{2}_{1}+a^{2}_{2}+a^{2}_{3}+a^{2}_{4}+a^{2}_{5}+a^{2}_{6}= \frac{45}{4}$. But I do not think this is the shortest of doing that, I am wondering if someone would be able to give me a hint so I can think better than this. Thank you

Answer 1

Partial answer: Consider the polynomial $$P\left(x\right)=x^{6}+\sum_{m=1}^{6}b_{m}x^{m-1} $$ where $b_{m}\in\mathbb{R} $ and assume that $a_{i},\, i=1,\dots,6 $ are the roots of this polynomial. By the Laguerre-Samuleson inequality we have that $$b_{1}=-\sum_{i=1}^{6}a_{i}=-\frac{15}{2} $$ $$b_{2}=\sum_{i=1}^{6}a_{i}^{2}=\frac{45}{4} $$ and $$-\frac{b_{1}}{2}+\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}\leq a_{i}\leq-\frac{b_{1}}{6}+\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}\sqrt{5}\tag{1} $$ ($(1)$ and the other inequalities obviously hold for every polynomial of degree $n$) so in our case we have $$0\leq a_{i}\leq\frac{5}{2} $$ but this not ensure the inequality. So we have to recall the Brunk's inequalities and Boyd-Hawkins inequalities. Let $$s=\frac{\sqrt{6b_{2}-b_{1}^{2}}}{2}. $$ If we assume that $$a_{1}\geq a_{2}\geq\dots\geq a_{6} $$ then $$-\frac{b_{1}}{6}+\frac{s}{\sqrt{5}}\leq a_{1}\leq-\frac{b_{1}}{6}+s\sqrt{5},\tag{2} $$ $$-\frac{b_{1}}{6}-s\sqrt{5}\leq a_{6}\leq-\frac{b_{1}}{6}-\frac{s}{\sqrt{5}}\tag{3} $$ and $$-\frac{b_{1}}{6}-s\sqrt{\frac{k-1}{6-k+1}}\leq a_{k}\leq-\frac{b_{1}}{6}+s\sqrt{\frac{6-k}{k}},\, k=2,\dots,5.\tag{4} $$ It is important to note that the equality holds in the RHS of $(2)$ if and only if the equality holds in the RHS of $(3)$ and this is equivalent to $$a_{1}=-\frac{b_{1}}{6}+s\sqrt{5} $$ $$a_{2}=\dots=a_{6}=-\frac{b_{1}}{6}-\frac{s}{\sqrt{5}}. $$ In this case the product is exactly $5/2$. In a similar way it is possible to prove that the equality in the RHS of $(4)$ holds if and only if$$a_{1}=\dots=a_{k} $$ $$a_{k+1}=\dots=a_{6} $$ hence $$a_{1}=\dots=a_{k}=-\frac{b_{1}}{6}+s\sqrt{\frac{6-k}{k}} $$ $$a_{k+1}=\dots=a_{6}=-\frac{b_{1}}{6}-s\sqrt{\frac{k}{6-k}} $$ and again if we maximize a root necessary the product is less or equal to $5/2$. This argument not consider all the possible cases but maybe it may be helpful for a complete proof.

Answer 2

Let $$b_i=a_i-\dfrac54,$$ then $$\sum_{i=1}^6b_i=\sum_{i=1}^6a_i-6\cdot\dfrac54 = 0,$$ $$\sum b_i^2=\sum_{i=1}^6a_i^2 - \dfrac52\sum_{i=1}^6a_i+\dfrac{25}{16}\cdot6 = \dfrac{15}8,$$ and we can seach for conditional maximum of $$\prod_{i=1}^6\left(b_i+\dfrac54\right).$$ Then we can use the Lagrange multipliers method.

For that, let search the greatest value of function $$f(\vec b, \lambda,\mu) = \prod_{i=1}^6\left(b_i+\dfrac54\right) + \lambda\sum_{i=1}^6b_i+\mu\left(\sum_{i=1}^6b_i^2-\dfrac{15}8\right),$$ the stationery points of which accords to zero derivatives $f'_{b_k}=0,\ f'_\lambda=0,\ f'_\mu=0,$ or $$\begin{cases} \left(b_k+\dfrac54\right)^{-1}\prod_{i=1}^6\left(b_i+\dfrac54\right) + \lambda + 2\mu b_k = 0\\[4pt] \sum_{i=1}^6\ b_i \ = \ 0\\[4pt] \sum_{i=1}^6\ b_i^2\ - \ \dfrac{15}8\ =\ 0\\[4pt] k=1\dots6, \end{cases}$$ $$\begin{cases} \prod_{i=1}^6\left(b_i+\dfrac54\right) + \lambda\left(b_k+\dfrac54\right) + 2\mu\left(b_k+\dfrac54\right)b_k = 0\\[4pt] \sum_{i=1}^6\ b_i \ = \ 0\\[4pt] \sum_{i=1}^6\ b_i^2\ = \ \dfrac{15}8\\[4pt] k=1\dots6. \end{cases}$$ Summation by $k$ for indexed equation gives $$6\prod_{i=1}^6\left(b_i+\dfrac54\right) + \lambda\dfrac{15}2 + \mu\dfrac{15}4 = 0,$$ or $$\prod_{i=1}^6\left(b_i+\dfrac54\right) + \lambda\dfrac54 + \mu\dfrac58 = 0,$$ so we have $$\begin{cases} \lambda b_k + 2\mu\left(b_k^2+\dfrac54b_k-\dfrac5{16}\right) = 0\\[4pt] \sum_{i=1}^6\ b_i \ = \ 0\\[4pt] \sum_{i=1}^6\ b_i^2\ = \ \dfrac{15}8\\[4pt] k=1\dots6. \end{cases}$$ Values of $\lambda$ and $\mu$ can be arbitrary, but the roots production still constant. So $$\begin{cases} (b_k=n)\vee(b_k=p)\\[4pt] np=-\dfrac5{16}\\[4pt] n<0,\ p>0\\[4pt] \sum_{i=1}^6\ b_i \ = \ 0\\[4pt] \sum_{i=1}^6\ b_i^2\ = \ \dfrac{15}8\\[4pt] k=1\dots6, \end{cases}$$ and for $b_1\leq b_2 \leq b_3 \leq b_4 \leq b_5 \leq b_6$ we have: $$\left[\begin{split} &b_1=b_2=b_3=b_4=b_5=b_6=n<0,\quad &6n=0,\quad &6n^2=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=b_4=b_5=n<0,\quad b_6=p=-\dfrac5{16n},\quad &5n=\dfrac5{16n},\quad &5n^2+p^2=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=b_4=n<0,\quad b_5=b_6=p=-\dfrac5{16n},\quad &4n=2\dfrac5{16n},\quad &4n^2+2p^2=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=n<0,\quad b_4=b_5=b_6=p=-\dfrac5{16n},\quad &3n=3\dfrac5{16n},\quad &3n^2+3p^2=\dfrac{15}8\\[4pt] &b_1=b_2=n<0,\quad b_3=b_4=b_5=b_6=p=-\dfrac5{16n},\quad &2n=4\dfrac5{16n},\quad &2n^2+4p^2=\dfrac{15}8\\[4pt] &b_1=n<0,\quad b_2=b_3=b_4=b_5=b_6=p=-\dfrac5{16n},\quad &n=5\dfrac5{16n},\quad &n^2+5p^2=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=b_4=b_5=b_6=p>0,\quad &6p=0,\quad &6p^2=\dfrac{15}8, \end{split}\right.$$ or $$\left[\begin{split} &b_1=b_2=b_3=b_4=b_5=b_6=n=0,\quad &6\cdot0=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=b_4=b_5=n=-\dfrac14<0,\quad b_6=p=\dfrac54,\quad &5\dfrac1{16}+\dfrac{25}{16}=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=b_4=n=-\sqrt{\dfrac5{32}},\quad b_5=b_6=p=\sqrt{\dfrac58},\quad &4\dfrac5{32}+2\dfrac58=\dfrac{15}8\\[4pt] &b_1=b_2=b_3=-\dfrac{\sqrt5}4,\quad b_4=b_5=b_6=\dfrac{\sqrt5}4,\quad &3\dfrac5{16}+3\dfrac5{16}=\dfrac{15}8\\[4pt] &b_1=b_2=n=-\dfrac{\sqrt5}2,\quad b_3=b_4=b_5=b_6=\dfrac{\sqrt5}8,\quad &2\dfrac54+4\dfrac5{64}=\dfrac{15}8\\[4pt] &b_1=n=-\dfrac54,\quad b_2=b_3=b_4=b_5=b_6=p=\dfrac14,\quad &\dfrac{25}{16}+5\dfrac1{16}=\dfrac{15}8\\ &b_1=b_2=b_3=b_4=b_5=b_6=p=0,\quad &6\cdot0=\dfrac{15}8, \end{split}\right.$$ Some systems are incompatible.

Therefore, the greatest value of goal production achieves in one of the points of the set $$\begin{pmatrix}b_1\\[10pt]b_2\\[10pt]b_3\\[10pt]b_4\\[10pt]b_5\\[10pt]b_6\end{pmatrix} \in\left\{ \begin{pmatrix}-\dfrac14\\-\dfrac14\\-\dfrac14\\-\dfrac14\\-\dfrac14\\\dfrac54\end{pmatrix}, \begin{pmatrix}-\sqrt{\dfrac5{32}}\\-\sqrt{\dfrac5{32}}\\-\sqrt{\dfrac5{32}}\\-\sqrt{\dfrac5{32}}\\\sqrt{\dfrac58}\\\sqrt{\dfrac58}\\\end{pmatrix}, \begin{pmatrix}-\dfrac{\sqrt5}4\\-\dfrac{\sqrt5}4\\-\dfrac{\sqrt5}4\\ \dfrac{\sqrt5}4\\\dfrac{\sqrt5}4\\\dfrac{\sqrt5}4\\\end{pmatrix}, \begin{pmatrix}-\dfrac54\\\dfrac14\\\dfrac14\\\dfrac14\\\dfrac14\\ \dfrac14\end{pmatrix} \right\}$$ and is equal to $$\max\left\{\left(-\dfrac14+\dfrac54\right)^5\left(\dfrac54+\dfrac54\right), \left(-\sqrt{\dfrac5{32}}+\dfrac54\right)^4\left(\sqrt{\dfrac58}+\dfrac54\right)^2, \left(-\dfrac{\sqrt5}4+\dfrac54\right)^3\left(\dfrac{\sqrt5}4+\dfrac54\right)^3, \left(-\dfrac54+\dfrac54\right)\left(\dfrac14+\dfrac54\right)^5\right\}$$ $$=\max\left\{\dfrac52,\dfrac{125(247+14\sqrt{10})}{16384} ,\dfrac{125}{64},0\right\}=\dfrac52.$$

That means that $$\boxed{\prod_{n=1}^6a_i \leq\dfrac52}$$

Answer 3

Here is a Calculus of Variations approach. Perhaps not terribly elegant, but it works. $$ \sum_{j=1}^6a_j=\frac{15}2\implies\sum_{j=1}^6\delta a_j=0\tag{1} $$ $$ \sum_{j=1}^6a_j^2=\frac{45}4\implies\sum_{j=1}^6a_j\delta a_j=0\tag{2} $$ To maximize $\prod\limits_{j=1}^6a_j$, we want $$ \sum_{j=1}^6\frac{\delta a_j}{a_j}=0\tag{3} $$ for all variations under the conditions $(1)$ and $(2)$. Linearity implies that there are constants $b$ and $c$ so that $$ \frac1{a_j}=b+ca_j\tag{4} $$ Multiplying $(4)$ by $a_k$ and shuffling yields $$ ca_j^2+ba_j-1=0\tag{5} $$ Equation $(5)$ implies that there are only two possible values for $a_j$, $h$ and $k$. All of the $a_j$ cannot be equal since then $(1)$ implies $a_j=\frac54$, which does not satisfy $(2)$.

Thus, we have $3$ cases: $$ h+5k=\frac{15}2\quad\text{and}\quad h^2+5k^2=\frac{45}4\implies(h,k)\in\left\{\left(0,\frac32\right),\left(\frac52,1\right)\right\}\tag{6} $$ This gives products of $0$ and $\frac52$. $$ 2h+4k=\frac{15}2\quad\text{and}\quad2h^2+4k^2=\frac{45}4\implies(h,k)=\left(\frac{5\pm\sqrt{10}}4,\frac{10\mp\sqrt{10}}8\right)\tag{7} $$ This gives products of $\frac{125(247\pm14\sqrt{10})}{16384}$. $$ 3h+3k=\frac{15}2\quad\text{and}\quad3h^2+3k^2=\frac{45}4\implies(h,k)=\left(\frac{5+\sqrt5}4,\frac{5-\sqrt5}4\right)\tag{8} $$ This gives a product of $\frac{125}{64}$.

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The greatest of these products is $\frac52$ given by the second solution in $(6)$: $\left\{\frac52,1,1,1,1,1\right\}$.

Therefore, $$ \prod_{j=1}^6a_j\le\frac52\tag{9} $$

Answer 4

Hint:

For any one of the variables, we have from CS inequality: $$5\left(\frac{45}4-a_i^2 \right) \geqslant \left(\frac{15}2-a_i \right)^2 \implies 0 \leqslant a_i \leqslant \frac52$$

If any $a_i = 0$ we clearly have a minimum, so all the variables are positive, and it is enough to show $\exists a, b \in \mathbb R$ s.t. $\forall x \in (0, \frac52]$. $$f(x) = \left(\tfrac16\log \tfrac52- \log x \right)+a(x-\tfrac54) + b(x^2 - \tfrac{15}8) \geqslant 0$$

A little investigation shows $a = \frac73 - \frac89\log \frac52, \quad b = -\frac23 + \frac49\log \frac52 $ fit the bill, and equality is iff $x \in \{1, \frac52\}$.

Answer 5

The feasible set $S$ is the transversal intersection of a $5$-sphere and a hyperplane in ${\mathbb R}^6$, hence a $4$-sphere, in particular: a smooth compact manifold. The function $p({\bf a}):=\prod_{i=1}^6 a_i$ assumes its maximum on $S$ in one or several conditionally stationary points of $p$. It follows that such maximum points will be brought to the fore using Lagrange's method. Therefore we set up the "Lagrangian" $$\Phi:=p-\lambda\sum_i a_i-\mu\sum_i a_i^2$$ and look at the equations $${\partial \Phi\over\partial a_i}={p\over a_i}-\lambda -2\mu a_i=0\qquad(1\leq i\leq 6)\ .$$ Since at the maximum points all $a_i$ have to be $\ne0$ this implies $$p-\lambda a_i-2\mu a_i^2=0\qquad(1\leq i\leq 6)\ .$$ This is saying that at the maximum points all $a_i$ are solutions of one and the same at most quadratic equation. It follows that at most two different values $a_i$ occur at such points. We therefore have to consider for $r\in\{0,1,2,3\}$ the equations $$r a_1+(6-r)a_2={15\over2},\qquad r a_1^2+(6-r)a_2^2={45\over4}\ ,$$ and have to check which case leads to the largest value of $p$. This amounts to the analysis conducted by @robjohn in his "variational" answer. I won't repeat it here.