Bài toán chi tiết
Cho a, b, c là các số thực thỏa mãn: \[ \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 5. \] Chứng minh rằng: \[ 15 \left( \frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{a^2} \right) - 2 \left( \frac{a^3}{b^3} + \frac{b^3}{c^3} + \frac{c^3}{a^3} \right) = 119 \]
| 1 cách giải | KiênĐC
\begin{align*} 15 \sum \frac{a^2}{b^2} - 2 \sum \frac{a^3}{b^3} &= 3 \cdot 5 \sum \frac{a^2}{b^2} - 2 \sum \frac{a^3}{b^3} \\[10pt] &= 3 \sum \frac{a^2}{b^2} - 2 \sum \frac{a^3}{b^3} \\[10pt] &= 3 \sum \frac{a b + \frac{b a^2}{c} + \frac{c a}{b}}{b^2} - 2 \sum \frac{a^3}{b^3} \\[10pt] &= \sum \frac{a b}{b^3} + 3 \sum \frac{a b + \sum \frac{c a}{b^2}}{c b} \\[10pt] &= \left( \sum \frac{a}{5} \right)^2 - 3 \sum \frac{a b}{b c} (a b c) \\[10pt] &\quad + 3 \sum \frac{a^2}{5c} + 3 \sum \frac{c a}{5^2} = \left\{ \sum x^3 = (5 x)^3 - 3 x (a + b c) - 6 x y z \right\} \\[10pt] &\approx 5^3 - 3 \sum \frac{a}{c (a b c)} \\[10pt] &= 125 - 3 = 119 \end{align*}
