$0=ab(\frac{c}{2}-\sqrt{c-1})+bc(\frac{a}{2}-\sqrt{c-1})+ca(\frac{b}{2}-\sqrt{c-1})$
$0=\frac{1}{2}(ab(\sqrt{c-1}-1)^2+bc(\sqrt{a-1}-1)^2)+ca(\frac{b-1}-1)^2))$
$事实上,由\sqrt{c-1},\sqrt{b-1},\sqrt{a-1}有意义知a,b,c\gt 0$
$故\sqrt{c-1}-1=\sqrt{b-1}-1=\sqrt{a-1}-1=0$
$即a=b=c=2$