Real solutions for $\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|$ Do there exist real numbers $a,b,c,d$ such that 
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}>|a-c|+|b-d|?$$

Consider the points $M=(a,b)$ and $N=(c,d)$. By the triangle inequality
$$OM-ON\leq MN \leq MP+NP $$
which gives us
$$\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\leq MN\leq |a-c|+|b-d| $$
Since $$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<\sqrt{a^2+b^2}-\sqrt{c^2+d^2} $$ we have the reverse inequality:
$$\sqrt{a^2+b^2}-\sqrt{1+c^2+d^2}<|a-c|+|b-d| $$

