Problem: $$x/2 - 4/x - 1 > 0$$
Simplified to: $$(x^2-8)/2x > 1$$
Right solution (put all at one side, bring to one fraction) is union of: $$x^2-8-2x>0, x > 0$$ and $$x^2-8-2x<0, x < 0$$
Wrong solution (multiply both sides by $2x$): $$x^2 - 8 > 2x$$
How formally explain why multiplying both sides by 2x is wrong here?
Set $x=-1$ in the inequality $$ \frac{x^2-8}{2x}>1 $$ You can check that it is valid. Now multiply by $2x$ to "get" $$ x^2-8>2x $$ which is invalid for the same value $x=-1$. The problem is that multiplication of inequality by negative number changes the sign of inequality. Formally $$ (a>b)\wedge (c<0)\implies ac<bc $$
