Is there a value of $t$ above which these inequalities aren't jointly satisfiable Let $\{a_{1}, ...., a_{8}\} \in [0, 1]$ be such that $\sum \limits_{0 < i \leq 8} a_{i} = 1$  and let $t \in (\frac{1}{2}, 1]$. Is there a value of $t$ above which the following inequalities aren't jointly satisfiable? I know that they are all satisfiable for any value of $t \leq \frac{2}{3}$. 
(1) $\frac{a_{1} + a_{2}}{a_{1} + a_{2} + a_{3} + a_{4}} \geq t$
(2) $\frac{a_{1} + a_{5}}{a_{1} + a_{2} + a_{5} + a_{6}} \geq t$
(3) $\frac{a_{4} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \geq t$
(4) $\frac{a_{7} + a_{8}}{a_{3} + a_{4} + a_{7} + a_{8}} \geq t$
(5) $\frac{a_{1} + a_{3}}{a_{1} + a_{2} + a_{3} + a_{4}} \leq 1 - t$
(6) $\frac{a_{6} + a_{8}}{a_{2} + a_{4} + a_{6} + a_{8}} \leq 1 - t$

Let’s start a quest for a largest $t<1$ for which there exist feasible $a_1,\dots, a_8$ that satisfy all six inequalities. In this quest we can drop the condition $\sum a_i=1$, because if non-negative numbers $a_i$ satisfy the inequalities for some $t$ then numbers $a’_i=\frac {a_i}{\sum a_j}$ also  satisfy the inequalities for the same $t$ and $\sum a’_i=1$. 
Now note that each of $a_5$ and $a_7$ occur in only one inequality, (2)  and (4), respectively. Pick any of these $a_i$’s and fix all other $a_j$’s. Then when $a_i$ tends to infinity, the left hand side of the inequality, in which it occurs, tends to $1$. Since $t<1$, for all other $a_j$’s the inequality will be satisfied for sufficiently large $a_i$. Thus in our quest we can drop both $a_5$ and $a_7$ and the inequalities (2)  and (4).
After routine transformations, we can simplify the remaining inequalities to the following
(1’) $a_1+a_2\ge s(a_3+a_4)$ 
(3’) $a_4+a_8\ge s(a_2+a_6)$
(5’) $a_2+a_4\ge s(a_1+a_3)$
(6’) $a_2+a_4\ge s(a_6+a_8)$.
Here $s=\frac {t}{1-t}>0,$ because $t>\frac 12$.
Since $a_3$ and $a_6$ occur only at the right hand sides of the inequalities, when we annulate them, the inequality still be  satisfied. 
Now the inequalities transform to  
(1’’) $a_1+a_2\ge sa_4$ 
(3’’) $a_4+a_8\ge sa_2$
(5’’) $a_2+a_4\ge sa_1$
(6’’) $a_2+a_4\ge sa_8$.
Adding (1’’) and (3’’), we obtain $a_1+a_8\ge (s-1)(a_2+a_4)$. Adding (5’’) and (6’’), we obtain $2(a_2+a_4)\ge s(a_1+a_8)$. Since, by   $1$, $a_1+a_2+a_3+a_4>0$, and $a_3=0$, we have that both  $A=a_1+a_8$ and $B=a_2+a_4$ are positive. Since $2A\ge 2(s-1)B\ge s(s-1)A$, we have $s(s-1)\le 2$. This implies $s\le 2$ and $t\le\frac 23$. 
In order to satisfy inequalities (1’’)-(6’’) with $s=2$ we may try to put $a_1=a_8=a$ and $a_2=a_4=b$. Then the inequalities will be satisfied iff $a=b$. If we put $a>0$, $a_3=a_6=0$,  $a_5=a_7=a$, the inequalities (2) and (4) also will be satisfied. At last, to achieve $\sum a_i=1$, we put $a=\frac 16$.
