Note that $A\left( k,k^{2} -1,1\right) =k^{2}$. Therefore, for any $k \geq 2$, $k$ does not have the property. We will show that $A( 1,a,b) \leq 2$. $A( 1,a,a) =\frac{2a}{a^{2}} =\frac{2}{a} \leq 2$. WLOG, $a< b$. $A( 1,a,b) =\frac{a+b}{a^{2} +b( b-a)} \leq \frac{a+b}{a^{2} +b} \leq \frac{a+b}{a+b} =1$. Hence the answer is $k=1$.