Lugging more logs

Given that $2^x3^y5^z = 30$ , $\implies x \ln 2 + y \ln 3 + z \ln 5 = \ln 30$ . By Cauchy-Schwarz inequality :

$\begin{aligned} (\ln^2 2 + \ln^2 3 + \ln^2 5) (x^2+y^2+z^2) & \ge (x \ln 2 + y \ln 3 + z \ln 5)^2 = \ln^2 30 \\ \implies x^2+y^2+z^2 & \ge \frac {\ln^2 30}{\ln^2 2 + \ln^2 3 + \ln^2 5} \approx \boxed{2.704} \end{aligned}$

Equality occurs when $\dfrac x{\ln 2} = \dfrac y{\ln 3} = \dfrac z{\ln 5}$ .

---

By Lagrange multiplier (Note that this is a calculus solution but the question is algebra.)

Given that Given that $2^x3^y5^z = 30$ , $\implies x \ln 2 + y \ln 3 + z \ln 5 = \ln 30$ . Then we have

$\begin{aligned} F(x,y,z,\lambda) & = x^2 + y^2 + z^2 - \lambda (x \ln 2 + y \ln 3 + z \ln 5 - \ln 30) \\ \frac {\partial F}{\partial x} & = 2x - \lambda \ln 2 \\ \frac {\partial F}{\partial y} & = 2y - \lambda \ln 3 \\ \frac {\partial F}{\partial z} & = 2z - \lambda \ln 5 \\ \frac {\partial F}{\partial \lambda} & = x \ln 2 + y \ln 3 + z \ln 5 - \ln 30 \end{aligned}$

Putting $\dfrac {\partial F}{\partial x} = \dfrac {\partial F}{\partial y} = \dfrac {\partial F}{\partial z} = 0$ , $\implies \dfrac x{\log 2} = \dfrac y{\log 3} = \dfrac z{\log 5} = \dfrac \lambda 2$ .

Putting $\dfrac {\partial F}{\partial \lambda} = 0$ , $\implies \dfrac \lambda 2 (\ln^2 2 + \ln^2 3 + \ln^2 5) = \ln 30$ .

And

$\begin{aligned} \min(x^2 + y^2 + z^2) & = \dfrac {\lambda^2}4 (\ln^2 2 + \ln^2 3 + \ln^2 5) \\ & = \dfrac {\ln^2 30}{\ln^2 2 + \ln^2 3 + \ln^2 5} \\ & \approx \boxed{2.704} \end{aligned}$

$2^x3^y5^z=30\implies (\ln 2)dx+(\ln 3)dy+(\ln 5)dz=0$

$x^2+y^2+z^2=$ minimum $\implies$

$xdx+ydy+zdz=0$

Using LaGrange's method of undetermined multipliers, we can write

$x=\lambda \ln 2,y=\lambda \ln 3,z=\lambda \ln 5$

So, $\lambda =\dfrac {\ln 30}{(\ln 2)^2+(\ln 3)^2+(\ln 5)^2}\approx 0.7951$

Hence $x\approx 0.551122,y\approx 0.8735,z\approx 1.27966$

Therefore the minimum value of $x^2+y^2+z^2$ is $\approx \boxed {2.704}$ .