very nice problem, i remember to have seen it in a "problems and solutions from around the world"... the part "i) implies ii)" consists on a very simple use of the cauchy schwarz inequality... and the "quadrilateral" inequality, if it can be called like that (a+b+c>d). by cauchy $(a^2x+b^2y+c^2z)\geq(a^2x+b^2y+c^2z)(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})\geq(a+b+c)^2>d^2$

Condition(ii) implies Condition(i): Let $ a = \frac{1}{x} $, $ b = \frac{1}{y} $, $ c = \frac{1}{z} $. Then $ d $ can be chosen such that $ 0 < (a + b + c) - d = (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) - d < e $, for all $ e > 0 $. Hence (assuming $ e < \frac{1}{x} + \frac{1}{y} + \frac{1}{z} $): $ a^{2}x + b^{2}y + c^2{z} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z} > d^{2} > (\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - e)^{2} $ Taking $ e \to 0 $: $ \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \geq (\frac{1}{x} + \frac{1}{y} + \frac{1}{z})^{2} \Leftrightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq 1 $

My solution: $(i)\rightarrow(ii)$: Since $ \dfrac 1x+\dfrac 1y+\dfrac 1z\leq1$ and $a+b+c>d$: $a^2x+b^2y+c^2z\ge \left(a^2x+b^2y+c^2z\right)\left(\dfrac 1x + \dfrac 1y + \dfrac 1z\right) \ge(a+b+c)^2>d^2$. We will prove $(ii)\rightarrow(i)$ by finding an example of $a,b,c,d$ when $(ii)$ doesn't hold if $(i)$ doesn't hold. Motivation: since we used CS we need $a^2x^2=b^2y^2=c^2z^2$ in order to get equality. It is equivivalent to choosing $a=\frac{1}{x}$ etc. Also, as our last step we used $a+b+c>d$ so it is logical to assume $d$ to be almost equal $a+b+c$. Finally, $a^2x+b^2y+c^2z=\dfrac 1x+\dfrac 1y+\dfrac 1z=t$. But $d=a+b+c-\epsilon= \dfrac 1x+\dfrac 1y+\dfrac 1z-\epsilon=t-\epsilon$. So we only need to find $\epsilon_t>0: t^2-2t\epsilon_t+\epsilon_t^2-t>0$, which is always possible since $t>1$. (take $0<\epsilon_t<t-\sqrt{t}$. It is possible because of $t>1\rightarrow t>\sqrt{t}$)