Ta có :
$ax^2=\frac{by^3}{x}=\frac{cz^3}{x}\Rightarrow 2ax^2=\frac{by^3}{x}+\frac{cz^3}{x}$
Tương tự ta có :
$2by^2=\frac{ax^3}{y}+\frac{cz^3}{y}$
$2cz^3= \frac{ax^3}{z}+\frac{by^3}{z}$
Cộng vế theo vế, ta có $2(ax^2+by^2+cz^2)=cz^3( \frac{1}{x} + \frac{1}{y})+by^3( \frac{1}{x}+\frac{1}{z})+ax^3( \frac{1}{y}+\frac{1}{z}) (*)$
Vì $ax^3=by^3=cz^3$ nên từ $(*)\Rightarrow 2(ax^2+by^2+cz^2)=ax^3( \frac{1}{x} + \frac{1}{y}+\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z})$
$\Rightarrow2(ax^2+by^2+cz^2)=2ax^3( \frac{1}{x}+\frac{1}{y} + \frac{1}{z})=2ax^3$
$\Rightarrow ax^2+by^2+cz^2=ax^3\Rightarrow \sqrt[3]{ax^2+by^2+cz^2}=x.\sqrt[3]{a}$
$\Rightarrow \frac{1}{x}.\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{a}$
Tương tự ta có
$ \frac{1}{y}.\sqrt[3]{ax^2+by^2+cz^2}= \sqrt[3]{b} ;\frac{1}{c}.\sqrt[3]{ax^2+by^2+cz^2}=\sqrt[3]{c}$
cộng vế theo vế $\Rightarrow \sqrt[3]{ax^2+by^2+cz^2}( \frac{1}{x}+ \frac{1}{y}+\frac{1}{z})=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$
$\Rightarrow$ đpcm