Ireland

First approach. Suppose that the LHS of (i) is false for some triple of positive numbers $a$, $b$, $c$ that satisfy (1), so that $1=2abc+ab+bc+ca$, but $ab+bc+ca<3/4$. Then, by the AM-GM inequality, $$\begin{array}{rl}abc& =\sqrt{(ab)(bc)(ca)}=(\sqrt[3]{(ab)(bc)(ca)}{)}^{3/2}\\ & \le {\left(\frac{ab+bc+ca}{3}\right)}^{3/2}<{\left(\frac{1}{4}\right)}^{3/2}=\frac{1}{8},\end{array}$$ whence $2abc+ab+bc+ca<\frac{2}{8}+\frac{3}{4}=1,$ in contradiction to our assumption. Hence, (i) holds, and (ii) now follows as an immediate consequence, because from (i) and (1) we see that $$2xyz=1-(xy+yz+zx)\le 1-\frac{3}{4}=\frac{1}{4}⟺xyz\le \frac{1}{8}.$$

Second approach. The proof of the LHS of (i) can be done directly as follows. By the AM-GM inequality, $$xyz={\left(\sqrt[3]{(xy)(yz)(zx)}\right)}^{3/2}\le {\left(\frac{xy+yz+zx}{3}\right)}^{3/2},$$ and so, with $t=(xy+yz+zx)/3$, we have that $$1=2xyz+xy+yz+zx\le 2t^{3/2}+3t⟺1\le 2(\sqrt{t}{)}^3+3(\sqrt{t}{)}^2,$$ i.e., letting $s=\sqrt{t}$, $r=2s$ $$1\le 2s^3+3s^2⟺4\le r^3+3r^2⟺0\le (r-1)(r+2{)}^2⟺r\ge 1.$$ Thus, $t\ge 1/4$, i.e., $xy+yz+zx\ge 3/4$, as claimed.

Here is a direct way to prove (ii). By the AM-GM inequality, $$xy+yz+zx\ge 3\sqrt[3]{(xy)(yz)(zx)}=3(xyz{)}^{2/3}=3s^2(\text{where }s=\sqrt[3]{xyz}).$$ Hence $1=2xyz+xy+yz+zx\ge 2s^3+3s^2$ and this is equivalent to $$0\ge 2(s^3+1)+3(s^2-1)=(s+1{)}^2(2s-1)⟺2s\le 1.$$ Equivalently, $xyz=s^3\le 1/8$. Again, (i) follows from this.

To prove (2), note that $$(x+y+z{)}^2\ge 3(xy+yz+zx)\ge \frac{9}{4}⟺x+y+z\ge \frac{3}{2},$$ with equality iff $$\frac{9}{4}=(x+y+z{)}^2=3(xy+yz+zx),$$ i.e., $x+y+z=3/2$, and $x=y=z$, i.e., $x=y=z=1/2$.