If $x>0$ so $y>0$, $z>0$. Hence, after assuming $x=tan\frac{\alpha}{2}$, $y=\tan\frac{\beta}{2}$ and $z=\tan\frac{\gamma}{2}$ we get $\alpha+\beta+\gamma=180^{\circ}$, and $\frac{3}{\sin\alpha}=\frac{4}{\sin\beta}=\frac{5}{\sin\gamma}$, which gives $\gamma=90^{\circ}$ and the rest is smooth.

Find all triples $(a,b,c)$ of real numbers satisfying the system of equations $\left\{\begin{matrix} 2(a+\frac{1}{a})=3(b+\frac{1}{b})=4(c+\frac{1}{c}),\\ ab+bc+ca=1.\end{matrix}\right.$ 7th grade Chinese Problem $\frac{a+\frac{1}{a}}{b+\frac{1}{b}}=\frac{3}{2}$ $\frac{3}{2}=\frac{(a^2+1)b}{(b^2+1)a}=\frac{(a+b)(a+c)b}{(b+a)(b+c)a}=\frac{(a+c)b}{(b+c)a}$ or $ab+3ac=2bc$ same way $\frac{(a+b)c}{(b+c)a}=2$ or $2ab+ac=bc$ So $ a=\frac{1}{\sqrt{15}}, b=\frac{5}{3\sqrt{15}},c=\frac{5}{\sqrt{15}}.$