$$f(t) = \log(t^2+5t)$$
You need the argument of the logarithm to be positive. Hence, you have
$$t^2+5t > 0$$
$$t(t+5) > 0$$
For a product to positive, either both factors are positive or both are negative. (This is an alternative to the other procedure posted.)
Case $1$:
$$t > 0 \quad t+5 > 0$$
$$t > 0 \quad t > -5$$
For $t > 0$, both conditions are true, so the factors are positive.
Case $2$:
$$t < 0 \quad t+5 < 0$$
$$t < 0 \quad t < -5$$
For $t < -5$, both conditions are true, so the factors are negative.
Hence, the domain becomes $(-\infty, -5)\cup(0, +\infty).$
First, the inequality you get by adding inequalities is implied by the inequalities you start with, but is not equivalent to them. That is, as you note, if $a\gt b$ and $c\gt d$, then $a+c\gt b+d$. But if $a+c\gt b+d$, you cannot then conclude that $a\gt b$ and $c\gt d$, or any other such combination. If it is the case that $x\geq 5$, and it is also the case that $x\geq 7$, then it is certainly the case that $2x\geq 12$, and hence that $x\geq 6$. This is not the best possible thing you can say about $x$, but it is nonetheless a true thing you can say about $x$ based on your assumptions.
That is, in general, adding inequalities will yield an inequality that is weaker than the conjunction of your original inequalities: any solution to the conjunction of the original inequalities will also be a solution to their sum, but not conversely.
Note that this is also true for equalities! If $x=3$ and $y=5$, then $x+y=8$; but from $x+y=8$ you cannot deduce that $x=3$ and $y=5$.
When you add inequalities (just as when you add equalities), you are simply obtaining a condition entailed by those inequalities. It's not as neat as taking the conjunction (intersection) or disjunction (union), because you do not have a bidirectional implication. Note that if $S_1$ is the solution to the first inequality, and $S_2$ is the solution to the second inequality, then
- $x\in S_1\cup S_2$ if and only if $x\in S_1$ or $x\in S_2$;
- $x\in S_1\cap S_2$ if and only if $x\in S_1$ and $x\in S_2$,
so you get information flowing in both directions. You don't get that with the sum (or product) of inequalities.
Second, to your system: you can't necessarily solve a system of inequalities by adding or subtracting, just like you can't always solve a system of equalities by adding and subtracting. It just may be that you can find a clever set of operations that will lead you to a conclusion that is narrow enough to identify $a$ and $b$ (just as with equalities), but then again it may not. However, that is fine: if from your assumptions you end up with the conclusion that you must have specific values of $a$ and $b$, then that means that if there is a solution, it must be the values of $a$ and $b$ that you found. But if not all steps you took are reversible, you must check that final determination in your original system. This, again, also occurs with equalities, as manipulation of equalities may lead to spurious solutions: from $x=1$, squaring you get $x^2=1$, which gives solutions $x=1$ and $x=-1$... but $x=-1$ does not solve the original equation. That's because the step of squaring is not "reversible" (it's an implication, not an "if and only if").
From $ab\gt 0$, you know that $a$ and $b$ have the same sign. If they are both negative, then $ab\gt a$ and $ab\gt b$ both hold, but then $a+b\lt 0$ does not. Thus, we conclude that $a$ and $b$ are both positive.
Then
$$\begin{align*}
ab\gt b &\iff ab-b\gt 0\\
&\iff b(a-1)\gt 0.\\
ab\gt a &\iff ab-a\gt 0\\
&\iff a(b-1)\gt 0.
\end{align*}$$
Since $b\gt 0$, then $b(a-1)\gt 0$ implies $a\gt 1$; and similalry, $a(b-1)\gt 0$ implies $b\gt 1$. So you end up with $a\gt 1$ and $b\gt 1$. Then you can verify that these all satisfy the original equations, so that's your solution.
Best Answer
Another way to think about it: Assuming $0<a\leq b$, divide the top and bottom of your fraction by $b$ to get
$$\frac{a}{\frac{a}{b}+1}.$$
$a$ is the smaller number and you're dividing it by a number greater than one, so the result is smaller than $a$.