Jtdylktuy

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’) $a_1+a_2ge s(a_3+a_4)$

(3’) $a_4+a_8ge s(a_2+a_6)$

(5’) $a_2+a_4ge s(a_1+a_3)$

(6’) $a_2+a_4ge s(a_6+a_8)$.

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’’) $a_1+a_2ge sa_4$

(3’’) $a_4+a_8ge sa_2$

(5’’) $a_2+a_4ge sa_1$

(6’’) $a_2+a_4ge sa_8$.

Adding (1’’) and (3’’), we obtain $a_1+a_8ge (s-1)(a_2+a_4)$. 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 $2Age 2(s-1)Bge s(s-1)A$, we have $s(s-1)le 2$. This implies $sle 2$ and $tlefrac 23$.

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 $a=frac 16$.

Let $a_1=a_4=frac{1}{2}$ and everything else equal $0$ (all fractions make sense for these choices since $a_1$ and $a_4$ appear in all denominators). Choose $t=1$. From $(1)$ we get $frac{1}{2}geq 1$ and from $(2)$ we get $1geq 1$, so they are not jointly satisfiable.

(Forgive me if I understood the problem wrong).