Ta có ${{V}_{1}}=\frac{1}{3}\pi {{b}^{2}}c,{{V}_{2}}=\frac{1}{3}\pi {{c}^{2}}b$

${{V}_{3}}=\frac{1}{3}\pi \cdot A{{H}^{2}}\cdot BH+\frac{1}{3}\pi \cdot A{{H}^{2}}\cdot CH$

và: $=\frac{1}{3}\pi \cdot A{{H}^{2}}\cdot BC=\frac{1}{3}\pi \cdot \frac{{{b}^{2}}{{c}^{2}}}{{{a}^{2}}}\cdot a=\frac{1}{3}\pi \frac{{{b}^{2}}{{c}^{2}}}{a}$

Do đó $\frac{1}{V_{3}^{2}}=\frac{1}{\frac{1}{3}\pi }.\frac{{{a}^{2}}}{{{b}^{4}}{{c}^{4}}}$

và $\frac{1}{V_{1}^{2}}+\frac{1}{V_{2}^{2}}=\frac{1}{\frac{1}{3}\pi }\left( \frac{1}{{{b}^{4}}{{c}^{2}}}+\frac{1}{{{b}^{2}}{{c}^{4}}} \right)$.

Vì tam giác ABC vuông tại A nên ${{a}^{2}}={{b}^{2}}+{{c}^{2}}$.

Mặt khác

$\frac{1}{{{b}^{4}}{{c}^{2}}}+\frac{1}{{{b}^{2}}{{c}^{4}}}=\frac{1}{{{b}^{2}}{{c}^{2}}}\left( \frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}} \right)=\frac{1}{{{b}^{2}}{{c}^{2}}}.\frac{{{b}^{2}}+{{c}^{2}}}{{{b}^{2}}{{c}^{2}}}=\frac{{{a}^{2}}}{{{b}^{4}}{{c}^{4}}}$Vậy $\frac{1}{V_{3}^{2}}=\frac{1}{V_{1}^{2}}+\frac{1}{V_{2}^{2}}$.