I know the answer of the current problem, but that doesn't mean I know how to calculate the answer. In fact, I want to know how to get the answer of the general version $(\frac{\lambda_1x+\lambda_2\sqrt{xy}+\lambda_3\sqrt[3]{xyz}}{x+y+z})_\max$.
When $\frac{x}{y} = \frac{y}{z} = 4$, the fraction reaches $\frac43$. And it's not hard to prove that \begin{align} x+y+z &= \frac34x+(\frac3{16}x+\frac34y)+(\frac1{16}x+\frac14y+z)\\ &\geq\frac34(x+\sqrt{xy}+\sqrt[3]{xyz})\\ &\Rightarrow f(x, y, z) \leq \frac43 \end{align} But how to know that $f(x,y,z)_{\max}=\frac43?$ To be more precisely, applying Lagrange multiplier method shows that $$1+\frac{\sqrt{xy}}{2x}+\frac{\sqrt[3]{xyz}}{3x} = \frac{\sqrt{xy}}{2y}+\frac{\sqrt[3]{xyz}}{3y} =\frac{\sqrt[3]{xyz}}{3z}$$ how to solve this equation?
We can solve this system by the following way.
Let $x=a^6$, $y=b^6$ and $z=c^6$, where $a$, $b$ and $c$ are positives.
Thus, we have
$$\frac{a^3}{2b^3}+\frac{a^2c^2}{3b^4}=\frac{a^2b^2}{3c^4}$$ and $$1+\frac{b^3}{2a^3}+\frac{b^2c^2}{3a^4}=\frac{a^2b^2}{3c^4}.$$ The first gives $$\frac{a}{2b^3}+\frac{c^2}{3b^4}=\frac{b^2}{3c^4}$$ and $$a=\frac{2b^5}{3c^4}-\frac{2c^2}{3b}=\frac{2(b^6-c^6)}{3bc^4}.$$ Let $b=\sqrt[6]tc$.
Thus, after substitution $a=\frac{2(t-1)c}{3\sqrt[6]t}$ in the second equation we obtain: $$1+\frac{\sqrt{t}}{2\left(\frac{2(t-1)}{3\sqrt[6]t}\right)^3}+\frac{\sqrt[3]t}{3\left(\frac{2(t-1)}{3\sqrt[6]t}\right)^4}=\frac{\left(\frac{2(t-1)}{3\sqrt[6]t}\right)^2\sqrt[3]t}{3}$$ or $$1+\frac{27t}{16(t-1)^3}+\frac{27t}{16(t-1)^4}=\frac{4(t-1)^2}{27}$$ or $$432(t-1)^4+729t^2=64(t-1)^6$$ or $$(t-4)(64t^5-128t^4+16t^3+512t^2-313t+92)=0$$ and since $$64t^5-128t^4+16t^3+512t^2-313t+92>0,$$ we obtain $$t=4$$ and $y=4z.$
Can you end it now?
We'll prove that $$64t^5-128t^4+16t^3+512t^2-313t+92>0.$$ Indeed, we'll replace $t$ on $\frac{t}{2}.$
Thus, we need to prove that $$4t^5-16t^4+4t^3+256t^2-313t+184>0.$$ Now, by AM-GM $$4t^5+\frac{1024}{27}t^2\geq3\sqrt[3]{(2t^5)^2\frac{1024}{27}t^2}=16t^4.$$ Thus, it's enough to prove that $$4t^3+\left(256-\frac{1024}{27}\right)t^2-313t+192>0,$$ which is true by AM-GM again: $$4t^3+\left(256-\frac{1024}{27}\right)t^2-313t+192>\left(256-\frac{1024}{27}\right)t^2+192-313t\geq$$ $$\geq2\sqrt{\left(256-\frac{1024}{27}\right)t^2\cdot192}-313t=\left(2\sqrt{\left(256-\frac{1024}{27}\right)\cdot192}-313\right)t>0.$$