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 

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.
