Aren't we allowed to subject $a <b$ or $a > b$ ?! If yes then assume both to be negative numbers we will have $ab > 0$ and $\frac{a}{b} > 0$ for $a< 0$ and $b<0$ Now $a+b < 0$ and $a - b < 0$ if $a > b$ Similarly we could argue upon other such cases

JBMO2020 wrote: Two of the numbers $a+b, a-b, ab, a/b$ are positive, the other two are negative. Find the sign of $b$ We have that $ab*\frac{a}{b}=a^2>0$ so $ab$ and $\frac{a}{b}$ have the same sign. $1)$ If $ab>0$, (i) if both $a,b$ positive the we should have that $a+b$ and $a-b$ are both negative. Then the product $(a+b)(a-b)=a^2-b^2>0\Longrightarrow a>b$Thats absurd cause $a-b<0$. (ii)If $a,b$ are both negative then we should have that $a^2-b^2>0\Longleftrightarrow a^2>b^2\Longleftrightarrow |a|>|b|$. $-|a|<b<|a|$ $a$ is negative so $a<b<-a$.that holds.So $b$ is negative. $2)$ If $ab<0$. then $a+b$ and $a-b$ are both positive. $a^2-b^2>0$.Like the previous statement we conclude to $-|a|<b<|a|$. $(i)$ $a<0$ and $b>0$ then $a<b<-a$.thats absurd cause $a-b<0$ and in the intial condition is $a-b>0$.so $b$ cant be positive. $(ii)$ now $a>0$ and $b<0$. from the $-|a|<b<|a|$. we have that $-a<b<a$.That holds. So in every case $b<0$