I don't have problems usually with these types of inequalities. They usually have a simple trick where you either take them two by two or else to get the question. Or factor out $x + y + z$ etc.
However when I tried to solve this it became kinda challenging.
I proved through AM-GM that : $$\frac{xy}{z} + \frac{yz}{x} + \frac{xz}{y} ≥ x + y + z $$
But this doesn't really help , as the condition is with ${x^2} + {y^2} + {z^2} = 1$.
I tried some other methods , but I failed. Any help is welcomed.
We observe that, if $x,y,z>0$ does not hold, then the global minimum does not exist . Therefore, we need the restriction $x,y,z>0$ .
Let $\thinspace\dfrac {xy}{z}=a,\thinspace \dfrac {xz}{y}=b,\thinspace \dfrac {yz}{x}=c$ . Then, the original problem is equivalent to :
$$\min \{a+b+c\}=X$$
where $ab+bc+ac=1$ and $a,b,c>0$ .
Since
$$ \begin{align}(a+b+c)^2&≥3(ab+bc+ac)=3\end{align}\tag 1 $$
We conclude that :
$$ \begin{align}|a+b+c|=a+b+c≥\sqrt 3\thinspace\thinspace\thinspace\thinspace\thinspace\tiny{\blacksquare}\end{align} $$
Indeed, since we need the condition $a+b+c>0$ here, this forces the condition $x,y,z>0$ to be satisfied .
$(1)\thinspace\thinspace\thinspace $ Note that, the inequality
$$ \small {\begin{align}\frac {(a-b)^2}{2}+\frac {(b-c)^2}{2}+\frac {(c-a)^2}{2}≥0\end{align}} $$
leads to :
$$a^2+b^2+c^2≥ab+bc+ac$$
which is equivalent to :
$$(a+b+c)^2≥3(ab+bc+ac)\thinspace .$$