3. Zadatak

Uvjeti zadatka su:

$$ax^3=by^3=cz^3,\tag{*}$$ $$\frac1x+\frac1y+\frac1z=1.\tag{**}$$

Sada imamo

$$\begin{aligned} A&=\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{\frac{ax^3}{x}+\frac{by^3}{y}+\frac{cz^3}{z}}\\ &\overset{(*)}=\sqrt[3]{ax^3\left(\frac1x+\frac1y+\frac1z\right)}\overset{(**)}=\sqrt[3]{ax^3}=x\sqrt[3]a. \end{aligned}$$

Zbog $(*)$ je i $A=y\sqrt[3]b=z\sqrt[3]c$, pa je

$$\sqrt[3]a=\frac Ax,\quad\sqrt[3]b=\frac Ay,\quad\sqrt[3]c=\frac Az,$$

odakle je

$$\sqrt[3]a+\sqrt[3]b+\sqrt[3]c=A\left(\frac1x+\frac1y+\frac1z\right)\overset{(**)}=A=\sqrt[3]{ax^2+by^2+cz^2},$$

tj.

$$\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]a+\sqrt[3]b+\sqrt[3]c.$$