My solution: $(a)$ Without loss of generality, assume $x_1=1$ So, $\frac{1}{x_2}+\frac{1}{x_3}>x_2+x_3\Longrightarrow x_2x_3<1<x_1x_2x_3=x_2x_3$ which is impossible. $(b)$ All three of them can't be lesser than $1$. Suppose all three of them are greater than $1$, we have $\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}<3<x_1+x_2+x_3$ which is false. I am not able to contradict the case when exactly two of them are less than $1$

we have that $\sum x_{1}x_{2}>\sum {\frac{1}{x_{1}}}$ hence $x_{1}x_{2}x_{3}=1$ hence $(1-x_{1})(1-x_{2})(1-x_{3})>0$ hence we have that proof

and with $x_{1}<1,x_{2}<1,x_{3}>1, x_{3}>1/(x_{1}x_{2}),1/x_{3}<x_{1}x_{2}$ then we have that $0<(1-x_{1}x_{2})(x_{1}-1)(1-x_{2})$ and that is not true

(a) If FTSOC one of them were $1$, WLOG let that value be $x_3$. Then $x_1x_2>1$ and $x_1+x_2<\tfrac{1}{x_1}+\tfrac{1}{x_2}=\tfrac{x_1+x_2}{x_1x_2}$, the latter of which rearranges as $x_1x_2<1$. Contradiction. $\blacksquare$ (b) If all were less than $1$, $x_1x_2x_3<1$, contradiction. If all were greater than $1$, $x_i>\tfrac{1}{x_i}$, and summing yields $x_1+x_2+x_3>\tfrac{1}{x_1}+\tfrac{1}{x_2}+\tfrac{1}{x_3}$, contradiction. If exactly two of these were less than $1$, WLOG let those values be $x_1$ and $x_2$. Since $x_3>\tfrac{1}{x_1x_2}$ from the first inequality, $x_1+x_2+\tfrac{1}{x_1x_2}<x_1+x_2+x_3$ and $\tfrac{1}{x_1}+\tfrac{1}{x_2}+\tfrac{1}{x_3}<\tfrac{1}{x_1}+\tfrac{1}{x_2}+x_1x_2$, so from the second inequality $$x_1+x_2+\frac{1}{x_1x_2}<x_1+x_2+x_3<\frac{1}{x_1}+\frac{1}{x_2}+\frac{1}{x_3}<\frac{1}{x_1}+\frac{1}{x_2}+x_1x_2.$$We rearrange as $\tfrac{1}{x_1x_2}-\tfrac{1}{x_1}-\tfrac{1}{x_2}<x_1x_2-x_1-x_2$, the LHS of which equals $\tfrac{1-x_1-x_2}{x_1x_2}$. Adding $1$ to both sides, $\tfrac{1-x_1-x_2+x_1x_2}{x_1x_2}<x_1x_2-x_1-x_2+1$, so $\tfrac{1}{x_1x_2}<1\iff x_1x_2>1$, contradiction. Hence, exactly one of these is less than $1$, which is the requested result. $\blacksquare$