Chứng minh rằng: Nếu 1/a + 1/b + 1/c = 2 và a + b + c = abc thì 1/a^2 + 1/b^2
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)