Curious Condition

We rearrange the condition to obtain $z=\dfrac{x+y}{1-xy}$ . This is exactly like the tangent sum formula, so we let $x=\tan\alpha$ , $y=\tan\beta$ , and therefore $z=\tan(\alpha+\beta)$ .

We rearrange the inequality to find $k\ge \dfrac{1}{xyz}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ . Thus we need to find the maximum value of $\dfrac{1}{xyz}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}$ . To do this, we minimize $x,y,z$ .

Since $x,y\ge \dfrac{\sqrt{3}}{3}$ , then $\alpha,\beta \ge 30^{\circ}$ . Since we want to minimize $x,y,z$ , we take $\alpha=\beta=30^{\circ}$ . Thus, $x=y=\tan 30^{\circ}=\dfrac{\sqrt{3}}{3}$ , and $z=\tan 60^{\circ} = \sqrt{3}$ .

Thus, $\dfrac{1}{xyz}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{10\sqrt{3}}{3}$ , so our answer is $10+3+3=\boxed{16}$ .

Sorry for the rushed solution, I created a much more thorough solution but I misclicked and all of it got deleted. :(

since it is clear that z>x,y......so for taking the minimum possible value of k let's assume that x=y=3/root over3=1/root over 3........now putting the value of x and y in xyz+x+y-z=0....we get the value of z=root over 3.....now putting the value of x,y and z in kxyz-xy-yz-xz greater than equal to 1.....we get k=(10*root over 3 )/3....therefore a+b+c=10+3+3=16.........