In trying to prove the following inequality: $0\leq ab\Longrightarrow (0\leq a\wedge 0\leq b)\vee(a\leq 0\wedge b\leq 0)$ the following proof by contradiction was tried
Proof: Let $0\leq ab$ and let,$\neg (0\leq a\wedge 0\leq b)$ and $\neg(a\leq 0\wedge b\leq 0)$. But $\neg(a\leq 0\wedge b\leq 0)\Longrightarrow \neg(a\leq 0)\vee\neg(b\leq 0)\Longrightarrow 0<a\vee o<b$. For $o<a\Longrightarrow\frac{1}{a}$. Also $0<a\Longrightarrow 0\leq a$. Thus $0\leq ab\wedge 0<\frac{1}{a}\Longrightarrow 0(\frac{1}{a})< ab(\frac{1}{a})\Longrightarrow 0<b\Longrightarrow 0\leq b$. Therefore $0\leq a\wedge 0\leq b$.
For $0<b$ we also can prove in the same way $0\leq a\wedge 0\leq b$. Hence $0\leq a\wedge 0\leq b$ a contradiction since we assumed $\neg(0\leq a\wedge 0\leq b)$. Thus we have $0\leq a\wedge 0\leq b$. But $(o\leq a\wedge 0\leq b)\Longrightarrow (0\leq a\wedge 0\leq b)\vee(a\leq 0\wedge b\leq 0)$
I don't understand what you're doing from the point you introduce $1/a$.
You've shown that $¬(a≤0∧b≤0)$ is equivalent to $a>0∨b>0$.
Similarly $¬(0≤a∧0≤b)$ is equivalent to $a<0∨b<0$.
So we're left with $¬(0≤a∧0≤b)$ and $¬(a≤0∧b≤0)$
is equivalent to:
$(a<0∨b<0)$ and $(a>0∨b>0)$. From here just show that $ab<0$, and you've obtained the contradiction.
EDIT: here's a full answer
$(a<0∨b<0) ∧ (a>0∨b>0) \\\iff (a<0∧a>0)∨(a<0∧b>0)∨(b<0∧a>0)∨(b<0∧b>0) \\\implies (a<0∧a\ge0)∨(ab<0)∨(ab<0)∨(b<0∧b\ge0) \\\iff ((a<0)∧\neg(a<0))∨(ab<0)∨((b<0∧\neg(b<0)) \\\iff F∨(ab<0)∨F \\\iff ab<0$