Chứng minh rằng: Nếu $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2$ và a + b + c = abc thì $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 2$.

Ta có: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 2$

$\Rightarrow {\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)^2} = 4$

$\Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2\left( {\frac{1}{{ab}} + \frac{1}{{bc}} + \frac{1}{{ac}}} \right) = 4$

$\Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2\left( {\frac{c}{{abc}} + \frac{a}{{abc}} + \frac{b}{{abc}}} \right) = 4$

$\Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2\frac{{a + b + c}}{{abc}} = 4$

$\Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} + 2\frac{{abc}}{{abc}} = 4$ (vì a + b + c = abc)

$\Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = 2$ (đpcm)