<s>[INST] 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 
 [/INST] 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,
$$
\prod_{j=1}^6a_j\le\frac52\tag{9}
$$
</s>