2. Zadatak

Jednakost

$$\frac{a}{1}+\frac{b}{2}+\frac{c}{3}=0$$

pišemo u obliku

$$2c+3b+6a=0.$$

Jednakost

$$\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$$

kvadriramo:

$$\frac{1}{a^2}+\frac{4}{b^2}+\frac{9}{c^2}+\frac{4}{ab}+\frac{6}{ac}+\frac{12}{bc}=1$$ $$\frac{1}{a^2}+\frac{4}{b^2}+\frac{9}{c^2}+\frac{4c+6b+12a}{abc}=1.$$

Kako je $4c+6b+12a=0$, slijedi

$$\frac{1}{a^2}+\frac{4}{b^2}+\frac{9}{c^2}=1.$$