Proof: If among $a+b-c, b+c-a, c+a-b$, one of the values is zero, the inequality (1) obviously holds.

Below, we prove the case when $a+b-c, b+c-a, c+a-b$ are all non-zero.
(1) If among $a+b-c, b+c-a, c+a-b$, only one is positive, let's assume $a+b-c>0$. In this case, there should be $b+c-a<0$ and $c+a-b<0$. Adding these two inequalities, we get $2a<0$, which contradicts $a+b-c>0$. Therefore, this situation is impossible.
(2) If among $a+b-c, b+c-a, c+a-b$, two are positive and the other is negative, in this case, inequality (1) obviously holds.
(3) If $a+b-c, b+c-a, c+a-b$ are all positive, then $a, b, c$ are the three side lengths of some $\triangle ABC$. Let the area of $\triangle ABC$ be $S$, noting the Heron's formula for the area of a triangle:
$S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{1}{2}(a+b+c)$.
Transforming, we get
$$4 S=\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$$

Thus, inequality (1) is equivalent to
$$a^{2}+b^{2} \geqslant 4 S$$

The proof of this inequality is straightforward. In fact, by the two-variable mean inequality, we have
$$a^{2}+b^{2} \geqslant 2 a b \geqslant 2 a b \sin C=4 S.$$

In summary, inequality (1) is proved.