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Well This Can Be easily solved By using Cauchy inequality in Titu's Lemma form. Click Here

Using Titu's Lemma,

$\begin{aligned} \frac{1}{1+a^2} + \frac{1}{1+b^2} + \frac{1}{1+ab} & = \frac{1^2}{1+a^2} + \frac{1^2}{1+b^2} + \frac{\sqrt{2}^2}{2+2ab} \\ &\leq \frac{(2+\sqrt{2})^2}{4+(a+b)^2} \\ &= \frac{\frac{(2+\sqrt{2})^2}{4}}{1+\frac{(a+b)^2}{4}} \end{aligned}$

So, $x \geq \frac{(2+\sqrt{2})^2}{4}$ and $z = 4$ . We can take $x = 3$ . Substituting this into the required expression yields $0$ .

First note that by AM-GM $ab\leq\frac{1}{2}$ and because $a^2+b^2=1$ we have $ab=\frac{1}{2}$ only when $a=b=\frac{1}{\sqrt(2)}$ . Using Titu's Lemma we can conclude that the LHS is $\geq \frac{9}{4+ab}\geq 2$ and so the minimum of the LHS is $2$ and it is archived when $a=b=\frac{1}{\sqrt(2)}$ . Now $\frac{x}{1+\frac{(a+b)^2}{z}}=\frac{xz}{z+1+2ab}$ and it is equal to the minimum when $ab=\frac{1}{2}$ and when $\frac{xz}{z+2}=2$ , i.e. when $xz=2z+4$ . Remembering that $x$ and $z$ are positive integers we have that the solutions are $(3,4)$ , $(4,2)$ , $(6,1)$ , but the only acceptable solution (for which the inequality hold for all $ab$ , see below) is when $z=4$ and $x=3$ and so $\lfloor\frac{x}{z}\rfloor=\lfloor\frac{3}{4}\rfloor=0$ .

The inequality $\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+ab}\geq \frac{3}{1+\frac{(a+b)^2}{4}}$ is oblivious: rearranging we can arrive at the following inequality $(2ab-1)(5(ab)^2+3(ab)+1)\leq 0$ which is true when $ab\leq\frac{1}{2}$ as in our case.

The other inequalities do not hold for all $0<ab\leq\frac{1}{2}$