${{{ \mbox{1). If denote: }|A_iA_j| = a_{ij} \mbox{; for } i,j\in\{1,2,3,4\} \mbox{ and } i\neq j \mbox{, then we: } \\ |GA_1|.|GA_1^{\prime}}| = |GA_2|.|GA_2^{\prime}}| = |GA_3|.|GA_3^{\prime}}| = |GA_4|.|GA_4^{\prime}}| = R^2 - |OG|^2 = \\ = \frac {1}{16}.(a_{12}^2 + a_{13}^2 + a_{13}^2 + a_{23}^2 + a_{24}^2 + a_{34}^2)\ \mbox{and: } \\ |GA_1|^2 = \frac {3.(a_{12}^2 + a_{13}^2 + a_{14}^2) - (a_{23}^2 + a_{24}^2 + a_{34}^2)}{16};\dots \\ \mbox{2). Now we: } |GA_1|.|GA_2|.|GA_3|.|GA_4|\le |GA_1^{\prime}|.|GA_2^{\prime}|.|GA_3^{\prime}|.|GA_4^{\prime}|\Leftrightarrow \\ \Leftrightarrow |GA_1|^2.|GA_2|^2.|GA_3|^2.|GA_4|^2\le (|GA_1|.|GA_1^{\prime}|).(|GA_2|.|GA_2^{\prime}|).(|GA_3|.|GA_3^{\prime}|).(|GA_4|.|GA_4^{\prime}|)\Leftrightarrow \\ \prod [{3.(a_{12}^2 + a_{13}^2 + a_{14}^2) - (a_{23}^2 + a_{24}^2 + a_{34}^2)}]\le (\sum_{1\le i < j\le 4}{a_{ij}^2})^4 \mbox{, but this is the AM - GM inequality.} \\ \mbox{3). We: } \frac {1}{|GA_1^\prime|} + \frac {1}{|GA_2^\prime|} + \frac {1}{|GA_3^\prime|} + \frac {1}{|GA_4^\prime|}\le \frac {1}{|GA_1|} + \frac {1}{|GA_2|} + \frac {1}{|GA_3|} + \frac {1}{|GA_4|}\Leftrightarrow \\ \Leftrightarrow\sum_{i = 1}^4{\frac {|GA_i|}{|GA_i|.|GA_i^\prime|}}\le\sum_{i = 1}^4{\frac {|GA_i^\prime|}{|GA_i|.|GA_i^\prime|} \Leftrightarrow |GA_1| + |GA_2| + |GA_3| + |GA_4|\le |GA_1^{\prime}| + |GA_2^{\prime}| + |GA_3^{\prime}| + |GA_4^{\prime}|.$ ${ \mbox{4). Using Cebasev's inequality, we: } \\ 4.\sum_{i = 1}^4{|GA_i|} \le (\sum_{i = 1}^4{|GA_i|^2}).(\sum_{i = 1}^4 \frac {1}{|GA_i|}) = \\ = \frac {1}{4}.(\sum_{1\le i < j\le 4}a_{ij}^2}).(\sum_{i = 1}^4 \frac {1}{|GA_i|}) = 4.\sum_{i = 1}^4{\frac {|GA_i|.|GA_i^\prime|}{|GA_i|}} = 4.\sum_{i = 1}^4{|GA_i^\prime|}.$

Let $A_1, A_2, A_3, A_4$ be denoted by $A, B, C, D$ and $A_1', A_2', A_3', A_4'$ by $A', B', C', D'$ respectively. Let $O$ be the circumcenter of the tetrahedron and $R$ the circumradius. Take vectors with origin $O$. Denote $OP$ by $\textbf{P}$ etc. Then $GA^2=(\textbf{G}-\textbf{A})^2=\textbf{G}^2+\textbf{A}^2-2\textbf{GA}=OG^2+R^2-2\textbf{GA}$. Adding the similar equations for $B, C, D$ we get $GA^2+GB^2+GC^2+GD^2=4OG^2+4R^2-2\textbf{G}(4\textbf{G})=4(R^2-OG^2)$. Hence, by AM-GM we have $(GA^2GB^2GC^2GD^2)\le (R^2-OG^2)^4$. But we also have $GA\cdot GA' = R^2-OG^2$ etc. So $(R^2-OG^2)^4=(GA GB GC GD)(GA' GB' GC' GD')$. Hence $GA\cdot GB\cdot GC\cdot GD\le GA'\cdot GB'\cdot GC'\cdot GD'$, which is the first inequality. $\Box$ Now, by Cauchy-Schwarz, $(GA+GB+GC+GD)^2\le (GA^2+GB^2+GC^2+BD^2)(1^2+1^2+1^2+1^2)$ and \[16=\left(\sum\sqrt{GA}\sqrt{\frac{1}{GA}}\right)^2\le(GA + GB + GC + GD)\left(\frac{1}{GA} + \frac{1}{GB} + \frac{1}{GC} + \frac{1}{GD}\right).\] All sums are positive, so multiplying the two inequalities we get \[\frac{1}{4}(GA^2+GB^2+GC^2+GD^2)\left(\frac{1}{GA}+\frac{1}{GB}+\frac{1}{GC}+\frac{1}{GD}\right)\ge (GA+GB+GC+GD).\] Hence \[\left(\frac{1}{GA}+\frac{1}{GB}+\frac{1}{GC}+\frac{1}{GD}\right)\ge (GA+GB+GC+GD)\frac{1}{R^2-OG^2}=\frac{1}{GA'}+\frac{1}{GB'}+\frac{1}{GC'}+\frac{1}{GD'}\] which is the second inequality. $\Box$

decade bump, here's a solution: Let $O$ and $R$ be the circumcenter and radius of the circumsphere of $A_{1}A_{2}A_{3}A_{4}$ respectively. We use Leibniz's theorem twice: \[4R^2 = \sum\limits_{i=1}^{4} OA_{i}^2 = \frac{1}{4}\sum\limits_{1\leq i<j\leq 4} A_{i}A_{j}^2 + 4\cdot GO^2 \Longrightarrow R^2 - GO^2 = \frac{1}{16}\sum\limits_{1\leq i<j\leq 4} A_{i}A_{j}^2\]\[\prod\limits_{i=1}^{4} GA_{i} \cdot GA_{i}^{'} = (R^2 - GO^2)^4 = \left(\frac{1}{16} \sum\limits_{1\leq i<j\leq 4} A_{i}A_{j}^2\right)^4 = \left(\frac{1}{4} \sum\limits_{i = 1}^{4} GA_{i}^2\right)^4 \geq \prod\limits_{i=1}^{4} GA_{i}^2\]by the AM-GM inequality. Therefore $\prod\limits_{i=1}^{4} GA_{i}^{'} \geq \prod\limits_{i=1}^{4} GA_{i}$ and we've proved the first inequality. For the second one, rewrite the left-hand expression: \[\sum\limits_{i=1}^{4} \frac{1}{GA_{i}^{'}} = \sum\limits_{i=1}^{4} \frac{GA_{i}}{R^2 - GO^2} = \frac{1}{R^2-GO^2}\sum\limits_{i=1}^{4} GA_{i} = \frac{4\sum\limits_{i=1}^{4} GA_{i}}{\sum\limits_{i=1}^{4} GA_{i}^2}\]Now, by the Chebyshev inequality, we have that: \[\left(\sum\limits_{i=1}^{4} GA_{i}^2\right)\left(\sum\limits_{i=1}^{4} \frac{1}{GA_{i}}\right) \geq 4\sum\limits_{i=1}^{4} GA_{i}\]hence the second inequality is also proven.

yeah so apparently $4\cdot \text{Pow}_G(S)=GA^2+GB^2+GC^2+GD^2$ (proof: LoC + projecting vertices onto $OG$, hopefully is right) yeah so the inequalities are really weak lmao :skull: 1) It suffices to show that \[(\text{Pow}_G(S))^4\ge GA^2\cdot GB^2\cdot GC^2\cdot GD^2\]or just that \[(\text{Pow}_G(S))^4\ge \left(\frac{1}{4}\left(GA^2+GB^2+GC^2+GD^2\right)\right)^4\]or that \[4\text{Pow}_G(S)\ge GA^2+GB^2+GC^2+GD^2\]which is an equality. 2) This is equivalent to \[GA'+GB'+GC'+GD'\ge GA+GB+GC+GD\]so it suffices to prove that \[(GA'+GB'+GC'+GD')(GA+GB+GC+GD)\ge (GA+GB+GC+GD)^2\]and we can weaken both sides: LHS by C-S and RHS by simple AM-GM. \[16\cdot \text{Pow}_G(S)\ge 4(GA^2+GB^2+GC^2+GD^2)\]which is an equality.