Set $m=a+b$ and $n=ab$. It boils down proving \[ |m^2-2n+n^2|\ge |mn+n| \]while $m^2\ge 4n$. Note that $m^2-2n+n^2\ge m^2/2+ n^2$, so $|m^2-2n+n^2| = m^2-2n+n^2$. Now, I prove $m^2-2n+n^2\ge mn+n$, the other case is similar. Notice this is equivalent to \[ \left(\frac{m}{2}-n\right)^2 + 3\frac{m^2-4n}{4}\ge 0, \]which is obvious. For equality, we must have $m/2 = n$ and $m^2=4n$. The second yields $a=b$. The first then gives $m/2 = a = n=a^2$ so $a(a-1)=0$, giving $(a,b)=(1,1)$.

Same as proving $(\frac{a}{b} + \frac{b}{a}+ab)^2 \geq (a+b+1)^2$ or $\frac{a^2}{b^2}+\frac{b^2}{a^2}+a^2b^2+2+2a^2+2b^2 \geq a^2+b^2+1+2ab+2a+2b$ or $\frac{a^2}{b^2}+\frac{b^2}{a^2}+a^2b^2+1+a^2+b^2 \geq 2ab+2a+2b$ But $\frac{a^2}{b^2}+\frac{b^2}{a^2}+a^2b^2+1+a^2+b^2 \geq a^2b^2+a^2+b^2+3 \geq 2ab+2a+2b$ with equality for $a=b=1$

JanHaj wrote: Show that for any real numbers $a$ and $b$ different from $0$, the inequality $$\bigg \lvert \frac{a}{b} + \frac{b}{a}+ab \bigg \lvert \geq \lvert a+b+1 \rvert$$holds. When is equality achieved? $$(\frac{a}{b}+\frac{b}{a}+ab)^2-(a+b+1)^2=(a-1)^2+(b-1)^2+(ab-1)^2+(\frac{a}{b}-\frac{b}{a})^2$$(战巡)