Unsolved Problem from Crux Solved

Problem

Unsolved Problem from Crux Solved

Solution 1

We use the method of Lagrange multipliers. Let

$\displaystyle\begin{align} L(a_1,a_2,...,a_6,\lambda,\mu)=&a_1a_2...a_6-\lambda\left(a_1+a_2+...+a_6 -\frac{15}{2}\right) \\ &-\mu\left(a_1^2+a_2^2+...+a_6^2-\frac{45}{4}\right). \end{align}$

Setting the partial derivatives to $0$, we get the original constraints and for all $i's$,

$\displaystyle a_1a_2...a_6-\lambda a_i -2\mu a_i^2 = 0$

for the critical points. Subtracting one such equation from another (for different $i's$),

$\displaystyle (a_i-a_j)(\lambda + 2\mu (a_i+a_j))=0.$

Thus, either $a_i=a_j$ or $a_i+a_j$ is constant. Moreover, whenever the sum is a constant, the constant does not depend on the values of $i$ and $j$. This implies the all the $a_i's$ can only take one of two possible values. Let the two values be $p$ and $q$. All the $a_i's$ have been partitioned into two equivalence classes with the classes identified by the value $a_i$ in a class takes. We can rule out the case of all $a_i's$ being in the same class by noting that the two constraints cannot be simultaneously satisfied in that case. So we are left with the following three combinations to look at: $(1,5)$, $(2,4)$, and $(3,3)$. The brackets indicate the number of $a_i's$ in each of the classes. Other cases are symmetrical.

Case 1:

$\displaystyle\begin{align} p+5q\,&=\frac{15}{2} \\ p^2+5q^2&=\frac{45}{4} \\ \end{align}$

Thus, $p=0, q=3/2$ and $\prod a_i=0$; or $p=5/2, q=1$ and $\prod a_i=pq^5=5/2$.

Case 2:

$\displaystyle\begin{align} 2p+4q\,&=\frac{15}{2} \\ 2p^2+4q^2&=\frac{45}{4} \\ \end{align}$

Thus, $p=(5-\sqrt{10})/4, q=(10+\sqrt{10})/8$ and $\prod a_i=p^2q^4\sim 1.547 \lt 5/2$; or $p=(5+\sqrt{10})/4, q=(10-\sqrt{10})/8$ and $\prod a_i=p^2q^4\sim 2.222 \lt 5/2$.

Case 3:

$\displaystyle\begin{align} p+q&=\frac{5}{2} \\ p^2+q^2&=\frac{15}{4} \\ \end{align}$

Thus, $p=(5-\sqrt{5})/4, q=(5+\sqrt{5})/4$ and $\prod a_i=p^3q^3=125/64 \lt 5/2$; or $p=(5+\sqrt{5})/4, q=(5-\sqrt{5})/4$ (same product).

All the critical points give products smaller than or equal to $5/2$. The critical point where the product has a maximum vale of $5/2$ is when one of the $a'i$ has value $5/2$ and all others have value $1$. Clearly, there are $6$ permutations embedded in this solution.

Solution 2

We have $\displaystyle a_1+a_2+a_3+a_4+a_5=\frac{15}{2}-a_6\,$ and $\displaystyle a_1^2+a_2^2+a_3^2+a_4^2+a_5^2=\frac{45}{4}-a_6^2.\,$ The function $g:\,\mathbb{R}\to\mathbb{R},\,$ $g(x)=x^2,\,$ is convex and, by Jensen's inequality,

$5(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2)\ge (a_1+a_2+a_3+a_4+a_5)^2,$

implying $\displaystyle 5\left(\frac{45}{6}-a_6^2\right)\ge\left(\frac{15}{2}-a_6\right)^2,\,$ so that $\displaystyle a_6^2-\frac{5}{2}a_6\le 0,\,$ i.e., $a_6\in\left[0,\frac{5}{2}\right].\,$ By analogy, $\{a_1,a_2,a_3,a_4,a_5,a_6\}\subset\left[0,\frac{5}{2}\right].$

If one of $a_i$'s is zero, the whole product vanishes and there's nothing to prove. We, thus, shall assume $\{a_1,a_2,a_3,a_4,a_5,a_6\}\subset\left(0,\frac{5}{2}\right].\,$ We have

$\displaystyle \sum_{1\le i\lt j\le 6}a_ia_j =\frac{1}{2}\left[\left(\sum_{i=1}^6a_i\right)^2-\sum_{i=1}^6a_i^2\right]$

so that $\displaystyle\sum_{1\le i\lt j\le 6}a_ia_j=\frac{45}{2}.\,$ We consider the polynomial $P:\,\mathbb{R}\to\mathbb{R},\,$ $P(x)=\displaystyle \prod_{i=1}^6(x-a_i).\,$ There are positive numbers $m,\,$ $n\,$ and $q\,$ such that

$\displaystyle P(x)=x^6-\frac{15}{2}x^5+\frac{45}{2}x^4-mx^3+nx^2-qx+p,$

where $p=a_1a_2a_3a_4a_5a_6.\,$ Now consider the function $f:\,(0,\infty)\to\mathbb{R},\,$ $\displaystyle f(x)=\frac{P(x)}{x}.\,$ Obviously, since $p\ne 0,\,$ the roots of $f\,$ are $a_i,\,$ $i=1,\ldots,6.\,$ By Rolle's theorem, the third derivative $f^{(3)}(x)\,$ of $f\,$ has at least three positive roots. But $\displaystyle f^{(3)}(x)=3\cdot\frac{20x^6-60x^5+45x^4-2p}{x^4},\,$ for $x\gt 0.$

Now, let $Q:\,(0,\infty)\to\mathbb{R},\,$ $Q(x)=20x^6-60x^5+45x^4-2p.\,$ Obviously, $f^{(3)}(x)=0\,$ iff $Q(x)=0.\,$ Let's study the variation of $Q(x)\,$ on $(0,\infty).

We have $Q'(x)=60x^3(x-1)(2x-3),\,$ which implies that $Q\,$ is strictly increasing on $(0,1],\,$ strictly decreasing on $\displaystyle \left[1,\frac{3}{2}\right],$ and again strictly increasing on $\displaystyle\left[\frac{3}{2},\infty\right).$

Observe that $Q(0_{+})=-2p\lt 0,\,$ $Q(\infty)=\infty\gt 0.\,$ Now,since $Q\,$ is known to have at least three positive roots, Rolle's theorem implies that $Q(1)\ge 0\,$ and $\displaystyle Q\left(\frac{3}{2}\right)\le 0.$

From $Q(1)\ge 0\,$ it follows that $\displaystyle p\le\frac{5}{2},\,$ i.e., $a_1a_2a_3a_4a_5a_6\le\displaystyle \frac{5}{2}.$

Note that, for $a_1=a_2=a_3=a_4=a_5=1\,$ and $\displaystyle a_6=\frac{5}{2},\,$ we have

$\displaystyle \begin{align} &a_1+a_2+a_3+a_4+a_5+a_6=\frac{15}{2},\\ &a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2=\frac{45}{4},\\ &a_1a_2a_3a_4a_5a_6=\frac{5}{2}. \end{align}$

Acknowledgment

Leo Giugiuc has posted this problem at the CutTheKnotMath facebook page with a comment to the effect that the problem has been published at the Crux Mathematicorum a year ago and during that time no solution has been submitted.

The above solution is due to Amit Itagi; Faryad D. Sahneh came up independently with practically the same solution.

Leo Giugiuc - the author of the problem - has kindly supplied his original solution (Solution 2).

 

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