Is this possible If $r_1, r_2, t_1,$ and $t_2$ are real numbers and if $\left|r_{1}\right|<\left|r_{2}\right|$, $\left|t_{1}\right|<\left|t_{2}\right|$
and 
$$
\left|\left(r_{1}-r_{2}\right)\left(t_{1}-t_{2}\right)\right|>\left|\left(r_{1}+r_{2}\right)\left(t_{1}+t_{2}\right)-2\left(r_{1}r_{2}+t_{1}t_{2}\right)\right|,
$$
does it hold that 
$
\left|\left(r_{1}+r_{2}\right)-\left(t_{1}+t_{2}\right)\right|\ge\left|\left|r_{1}\right|-\left|t_{1}\right|\right|+\left|\left|r_{2}\right|-\left|t_{2}\right|\right|?
$ Thanks in advance.

The second inequality $$|r_{1}+r_{2}-(t_{1}+t_{2})|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$holds if $r_{1}> t_{1}>0$, $r_{2}> 0> t_{2}$, then we have $$|r_{1}+r_{2}-(t_{1}+t_{2})|=|r_{1}-t_{1}|+|r_{2}-t_{2}|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$
By assumption $r_{1}<r_{2}$, $0<t_{1}<-t_{2}$, then the left hand of the first inequality become $$(r_{2}-r_{1})(t_{1}-t_{2})=r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}$$
The right hand side now is depended on whether $$-2t_{1}t_{2}<2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})$$
If this is true when we want to have $$r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}>2r_{1}r_{2}+2t_{1}t_{2}-t_{1}r_{1}-t_{2}r_{2}-t_{1}r_{2}-t_{2}r_{1}$$
Re-arranging the terms we need to have $r_{2}t_{1}+r_{1}t_{2}>r_{1}r_{2}+t_{1}t_{2}$.
Let $r_{2}=50,r_{1}=1,t_{2}=-4,t_{1}=3$, then left hand become $150-4=146$, right hand become $50-12=38$. While $-2t_{1}t_{2}=24$,$2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})=100+50=150$.
Now go back to the original inequality we have:
$$49*7=343>|51*-1-2*(50-12)|=127$$ and $$|51-(-1)|=52>48=2+46$$
