Note that everything is invariant if we change all the signs, so w.l.o.g. $a>0$. Moreover, if $b$ and $c$ have the same sign, then $\vert bc+1\vert>1$ already. Since we also have the symmetry $b \leftrightarrow c$, we may thus w.l.o.g. assume that $b>0$ and $c<0$. Replacing $c$ by $-c$, it thus suffices to prove that \[\frac{b}{a}+\frac{b}{c}+\left\vert \frac{c}{a}-\frac{c}{b}\right\vert>1\]for $a,b,c>0$ (it is clear that we can now ignore the last term by homogeneity). If $b>a$ or $b>c$, this is obvious. Now assume that $b \le a$ and $b \le c$. In particular $\frac{c}{a}-\frac{c}{b} \le 0$ so the LHS becomes \[\frac{b}{a}+\frac{b}{c}+\frac{c}{b}-\frac{c}{a}.\]If $c<a$, then $\frac{c}{a}<1$ and $\frac{b}{c}+\frac{c}{b} \ge 2$ and we are done. So now assume that $b \le a \le c$. But now it suffices to prove that \[b^2(a+c)+c^2(b-a)>abc\]which is trivial after writing $a=b+d, c=b+d+e$ and expanding.

a_507_bc wrote: If $a, b, c$ are non-zero reals, prove that $|\frac{b} {a}-\frac{b} {c}|+|\frac{c} {a}-\frac{c}{b}|+|bc+1|>1$.

Attachments:

Look at $|bc+1|$; If $bc>0$ or $bc<-2$, then $|bc+1|>1$ and we are done. So assume that $-2 \leq bc <0$. Thus, one of $b$ or $c$ is positive and the other is negative. The inequality is symmetric in $b$ and $c$, so we can assume that $c<0<b$. Considering $-c=c'>0$, we should prove that $$|\frac{b} {a}+\frac{b} {c'}|+|\frac{c'}{b}-\frac{c'} {a}|+|1-bc'|>1$$where $0<bc'\leq2$ and $b$,$c>0$. Clearly we can assume that $a>0$ (why?). So the inequality is equivalent to $$\frac{b} {a}+\frac{b} {c'}+|\frac{c'}{b}-\frac{c'} {a}|+|1-bc'|>1.$$If $b\geq a$ or $b\geq c'$, there is nothing to prove. So assume that $b<a,c'$. Thus $\frac{c'}{b}>\frac{c'} {a}$ and we need to show that $$\frac{b} {a}+\frac{b} {c'}+\frac{c'}{b}-\frac{c'} {a}+|1-bc'|>1 \iff \frac{b} {a}+\frac{b} {c'}+\frac{c'}{b}+|1-bc'|>1 + \frac{c'} {a}.$$if $a \geq c'$, then $\frac{b} {c'}+\frac{c'}{b} \geq 2 \geq 1 + \frac{c'} {a}$ and we are done. Finally we consider the case $a<c'$. We have $$ \frac{b} {a}+\frac{b} {c'}+\frac{c'}{b}+|1-bc'|>1 + \frac{c'} {a} \iff \frac{b} {a}+\frac{a}{b}+\frac{b} {c'}+\frac{c'-a}{b}+|1-bc'|>2 + \frac{c'-a} {a}.$$And this is true. Because $\frac{b} {a}+\frac{a}{b} \geq 2$ and by previous assumptions $\frac{c'-a}{b} > \frac{c'-a} {a}$.$\blacksquare$

Change $\frac{1}{a}$ to $a$. Then the LHS is just some linear functions on some intervals, it attains its minimum at the endpoints of intervals or at the infinity, so we only need to check $a=\pm \infty, b,c$. When $a=\infty$ the inequality is obvious, when $a=c$ we get $|1-\frac{c}{b}|+|bc+1|>1$. Note that at least one of the modulos is strictly bigger than one depending on the sign of $bc$, $a=b$ can be done similarly