Question

CMR:neu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=R\) va a+b+c=abc thi \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

Answer 1

Sửa đề: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

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

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

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ac}=4\)

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

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

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

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