$\frac{(x^2+y)(y^2+x)}{(x+y-1)^2}\geq x+y-\frac{1}{4}$ it's $(2v^2-6u+1)^2\geq0$, where $x+y=2u$ and $xy=v^2$.

Hint. We have $$(x^2+y)(x+y^2)-(x+y-\frac{1}{4})(x+y-1)^2 = \frac{1}{4}(x^2(2y-3)^2-2(2y-3)(3y-1)x+(3y-1)^2) \ge 0$$

Note that $x^2+y=(x+y-1)+(x^2-x+1)$ and $x+y^2=(x+y-1)+(y^2-y+1)$. Hence, \[ (x^2+y)(x+y^2)=(x+y-1)^2+[(x^2-x+1)+(y^2-y+1)](x+y-1)+(x^2-x+1)(y^2-y+1). \]Moreover, note that if $a=x-\frac12$ and $b=y-\frac12$, then $a+b=x+y-1$ and \[ (x^2-x+1)+(y^2-y+1)=a^2+b^2+\frac34\cdot 2=(a+b)^2-2\left(ab-\frac34\right), \]\[ (x^2-x+1)(y^2-y+1)=(a^2+\frac34)(b^2+\frac34)=(ab)^2+\frac34(a^2+b^2)+\frac9{16}=(ab-\frac34)^2+\frac34(a+b)^2. \]It follows that if $t=ab-\frac34$, then \[ (x^2+y)(x+y^2)=(a+b)^2+[(a+b)^2-2t](a+b)+t^2+\frac34(a+b)^2, \]and thus \[ \frac{(x^2+y)(x+y^2)}{(x+y-1)^2}=\frac74+a+b-\frac{2t}{a+b}+\frac{t^2}{(a+b)^2} \]\[ =a+b+\left(\frac{t}{a+b}-1\right)^2+\frac34 \]\[ =x+y-\frac14+\left(\frac{ab-\frac34}{a+b}-1\right)^2\ge x+y-\frac14, \]and summing these cyclically yields the inequality. For the equality case, we need $ab-\frac34=a+b$ cyclically, and hence $b=f(a)$ etc. where $f(u)=\frac{u+\frac34}{u-1}=1+\frac{7/4}{u-1}$. We have $f(u)=u$ when $u=1+\frac{7/4}{u-1}$, $(u-1)^2=7/4$, $u_{1,2}=1\pm \sqrt7/2$. Note that $u-1$ and $f(u)-1$ have the same sign, and thus the sequence $u,f(u),f(f(u)),\dots$ are all in $I_1=(1;\infty)$ or all in $I_2=(-\infty;1)$. Suppose the sequence is in $I_j$. Then since $f$ is decreasing in $I_j$, $u>u_j$ implies $f(u)<f(u_j)=u_j$ and so $f(f(u))>u_j$, $f(f(f(u)))<u_j<u$. Similarly, $f(f(f(u)))\neq u$ for $u<u_j$ as well. Hence, $f(f(f(u)))=u$ if and only if $u\in\{u_1,u_2\}$. Since $f(f(f(a)))=a$ if equality holds, we need $a=1\pm \sqrt7/2$, i.e. $x=3/2\pm \sqrt7/2$. Conversely, if $x=y=z=3/2\pm \sqrt7/2$, equality indeed holds, as our estimates have equality and $x+y,y+z,z+x\neq 1$. Therefore the criterion for equality is \[ \boxed{x=y=z=\frac{3\pm \sqrt 7}{2} }. \]