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}$

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

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 $x \in \{1, \frac52\}$.