Problem 1576

Proof: Let $\boldsymbol{u}=(a, b), \boldsymbol{v}=(c, d)$, then the inequality
$$\sqrt{(a+c)^{2}+(b+d)^{2}} \leqslant \sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}$$

is a triangle inequality constructed by vectors $\boldsymbol{u}$ and $\boldsymbol{v}$, which is obviously true. Therefore, to prove the original inequality, it suffices to prove
$$\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}} \leqslant \sqrt{(a+c)^{2}+(b+d)^{2}}+\frac{2|a d-b c|}{\sqrt{(a+c)^{2}+(b+d)^{2}}}$$

which is equivalent to proving
$$\left(\sqrt{a^{2}+b^{2}}+\sqrt{c^{2}+d^{2}}\right)^{2} \leqslant(a+c)^{2}+(b+d)^{2}+2|a d-b c| .$$

And (1) is equivalent to $a^{2}+b^{2}+c^{2}+d^{2}+2 \sqrt{a^{2}+b^{2}} \cdot \sqrt{c^{2}+d^{2}}$
$$\begin{aligned} & \leqslant a^{2}+2 a c+c^{2}+b^{2}+2 b d+d^{2}+2|a d-b c| \\ \Leftrightarrow & \sqrt{a^{2}+b^{2}} \cdot \sqrt{c^{2}+d^{2}} \leqslant a c+b d+|a d-b c| \\ \Leftrightarrow & \left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right) \leqslant a^{2} c^{2}+2 a b c d+b^{2} d^{2}+a^{2} d^{2}-2 a b c d \\ & +b^{2} c^{2}+2(a c+b d)|a d-b c| \\ \Leftrightarrow & 0 \leqslant 2(a c+b d)|a d-b c| . \end{aligned}$$

The last inequality is obviously true, hence the original inequality holds.