Let $a=x^2$, $b=y^2$, $c=z^2$, and $d=t^2$ for $x,y,z,t>0$. From the given conditions, we have $x>z$, $y<t$, $x^2+y \ge z^2+t$, and $x+y^2\le z+t^2$. In particular, $x^2-z^2 \ge t-y$ and $t^2-y^2\ge x-z$. Multiplying these inequalities, while keeping in mind $x-z>0$ and $t-y>0$, we obtain $(x+z)(y+t)\ge 1$. Now, assume the hypothesis is false, and $x^2+y^2+z^2+t^2 \le 1$. Then $2(x^2+z^2)> (x+z)^2$, $2(y^2+t^2)>(y+t)^2$ by Cauchy-Schwarz (note that these inequalities are strict as $x>z$ and $y>t$). Thus $2>(x+z)^2 + (y+t)^2 \ge 2(x+z)(y+t)$, implying $1>(x+z)(y+t)$, contradiction with the earlier finding. This concludes the proof.

Note that $a-c=(\sqrt{a}+\sqrt{c})(\sqrt{a}-\sqrt{c})$ and $d-b=(\sqrt{d}+\sqrt{b})(\sqrt{d}-\sqrt{b})$ Therefore the first condition implies that $(\sqrt{a}+\sqrt{c})(\sqrt{a}-\sqrt{c})\geq \sqrt{d}-\sqrt{b}$. Multipliying this by $\sqrt{d}+sqrt{b}$ and we get $(\sqrt{a}+\sqrt{c})(\sqrt{a}-\sqrt{c})(\sqrt{d}+\sqrt{b})\geq (\sqrt{d}-\sqrt{b})(\sqrt{d}+\sqrt{b})\geq \sqrt{a}-\sqrt{c} (\text{by the second condition})$ Then $\sqrt{ab}+\sqrt{ad}+\sqrt{bc}+\sqrt{cd}=(\sqrt{a}+\sqrt{c})(\sqrt{b}+\sqrt{d})\geq 1$ and by $MA-MG$ we have $a+b\geq 2\sqrt{ab}$ and analogous with the others $\sqrt{}$, sum all and we get $2(a+b+c+d)\geq 2(\sqrt{ab}+\sqrt{ad}+\sqrt{bc}+\sqrt{cd})\geq 2$ then $a+b+c+d\geq 1$ and equality happens when $a=b=c=d$ and it is a contradiction for the hypothesis, therefore $a+b+c+d>1$

plagueis wrote: Let $a$, $b$, $c$ and $d$ positive real numbers with $a > c$ and $b < d$. Assume that \[a + \sqrt{b} \ge c + \sqrt{d} \qquad \text{and} \qquad \sqrt{a} + b \le \sqrt{c} + d\]Prove that $a + b + c + d > 1$. Proposed by Victor Domínguez $$a + b + c + d > \frac{(\sqrt a+\sqrt c)^2}{2}+\frac{(\sqrt b+\sqrt d)^2}{2}\geq (\sqrt a+\sqrt c)(\sqrt b+\sqrt d)\geq 1$$