We can consider the case $a \geq b \geq c$ which implies $c \leq 1$. The given inequality writes

$$a+b-2 \geq (ab-1)c \geq (ab-1)c^2 = (ab-1) \frac{3-a^2b^2}{a^2+b^2}$$

Put $x = \sqrt{ab}$. From the inequality $3a^2b^2 \geq a^2b^2 + b^2c^2 + c^2a^2 = 3$ we infer $x \geq 1$ and from $a^2b^2 < a^2b^2 + b^2c^2 + c^2a^2 = 3$ we find $x \leq \sqrt[4]{3}$. As $a+b \geq 2\sqrt{ab} = 2x$ and $a^2 + b^2 \geq 2ab = 2x^2$, to prove the inequality (1) it will suffice to show that

$$2(x-1) \geq (x^2-1) \frac{3-x^4}{2x^2}$$

As $x-1 \geq 0$, the last inequality is equivalent to

$$4x^2 \geq (x+1)(3-x^4)$$

which can be easily obtained by multiplying the obvious ones $2x^2 \geq x+1$ and $2 \geq 3-x^4$.

Equality holds only in the case when $a=b=c=1$.

Comment: As it is, the solution is incorrect, it only proves the weaker inequality $a+b-2 \geq (ab-1)c^2$, that is: $a+b+c^2 \geq abc^2 + 2$. The problem committee could not find a reasonable solution. Instead the problem could be slightly modified so that the method of the proposed solution applies. The modified problem is: