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}$ 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} $.

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 
 A: 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 $\frac{125}{64}$.

The greatest of these products is $\frac52$ given by the second solution in $(6)$: $\left\{\frac52,1,1,1,1,1\right\}$.
Therefore,
$$
\prod_{j=1}^6a_j\le\frac52\tag{9}
$$
A: 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 $x \in \{1, \frac52\}$.  
A: 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.
A: 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.
