The given condition leads to $$x-z=\sqrt{y+\sqrt{z}}-\sqrt{y+\sqrt{z}}$$ $$(x-z)\Big(\sqrt{y+\sqrt{z}}+\sqrt{y+\sqrt{z}}\Big)=\Big(\sqrt{y+\sqrt{z}}-\sqrt{y+\sqrt{z}}\Big)\Big(\sqrt{y+\sqrt{z}}+\sqrt{y+\sqrt{z}}\Big)=\sqrt{x}-\sqrt{z}$$ Because $x\ne z$, dividing both sides by $(\sqrt{x}-\sqrt{z})$ gives $$(\sqrt{x}+\sqrt{z})\Big(\sqrt{y+\sqrt{z}}+\sqrt{y+\sqrt{z}}\Big)=1$$ On the other hand, because $x, y, z$ are all positive, we have $\sqrt[4]{x}<\sqrt{y+\sqrt{x}}$ and $\sqrt[4]{z}<\sqrt{y+\sqrt{z}}$. Therefore $$(\sqrt{x}+\sqrt{z})\Big(\sqrt[4]{x}+\sqrt[4]{z}\Big) < (\sqrt{x}+\sqrt{z})\Big(\sqrt{y+\sqrt{z}}+\sqrt{y+\sqrt{z}}\Big)=1$$ Apply AM-GM: $2\sqrt[4]{xz} \le \sqrt{x}+\sqrt{z}$ and $2\sqrt[8]{xz} \le \sqrt[4]{x} + \sqrt[4]{z}$. Multiplying these two gives $$4\sqrt[8]{(xy)^3}\le(\sqrt{x}+\sqrt{z})(\sqrt[4]{x}+\sqrt[4]{z})<1$$ Thus $$(xz)^{3/8}< 4 \implies 40xz<1$$