\(\text{well.... if this is true for all $a,b,c,x,y,z$, then I can choose $(x,y,z)=2\sqrt{3}(1,1,1)$}\\ (a,b,c)\cdot (x,y,z)=30\\ 2\sqrt{3}(a+b+c)=30\\ (a+b+c) = 5\sqrt{3}\\ x+y+z = 6\sqrt{3}\\ \dfrac{a+b+c}{x+y+z} = \dfrac{5\sqrt{3}}{6\sqrt{3}}=\dfrac 5 6\)
+158 
