CMR: a^2/b^2 + b^2/c^2 + c^2/a^2 ≥ a/c + c/b + b/a 02/08/2021 Bởi Adeline CMR: a^2/b^2 + b^2/c^2 + c^2/a^2 ≥ a/c + c/b + b/a
Đáp án : $\left(\dfrac{a}{b} – \dfrac{b}{c}\right)^2 \geq 0 $ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} – 2\dfrac{a}{b}\cdot\dfrac{b}{c} \geq 0 $ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} \geq \dfrac{2a}{c} $ $\dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} \geq \dfrac{2b}{a} $ $\dfrac{c^2}{a^2} + \dfrac{a^2}{b^2} \geq \dfrac{2c}{b} $ $2\left(\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\right) \geq 2\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right) $ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\geq \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} $ Bình luận
Ta có: $\left(\dfrac{a}{b} – \dfrac{b}{c}\right)^2 \geq 0$ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} – 2\dfrac{a}{b}\cdot\dfrac{b}{c} \geq 0$ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} \geq \dfrac{2a}{c}$ Tương tự ta được: $\dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} \geq \dfrac{2b}{a}$ $\dfrac{c^2}{a^2} + \dfrac{a^2}{b^2} \geq \dfrac{2c}{b}$ Cộng vế theo vế ta được: $2\left(\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\right) \geq 2\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right)$ $\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\geq \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ Bình luận
Đáp án :
$\left(\dfrac{a}{b} – \dfrac{b}{c}\right)^2 \geq 0 $
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} – 2\dfrac{a}{b}\cdot\dfrac{b}{c} \geq 0 $
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} \geq \dfrac{2a}{c} $
$\dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} \geq \dfrac{2b}{a} $
$\dfrac{c^2}{a^2} + \dfrac{a^2}{b^2} \geq \dfrac{2c}{b} $
$2\left(\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\right) \geq 2\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right) $
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\geq \dfrac{a}{c} + \dfrac{c}{b} + \dfrac{b}{a} $
Ta có:
$\left(\dfrac{a}{b} – \dfrac{b}{c}\right)^2 \geq 0$
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} – 2\dfrac{a}{b}\cdot\dfrac{b}{c} \geq 0$
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} \geq \dfrac{2a}{c}$
Tương tự ta được:
$\dfrac{b^2}{c^2} + \dfrac{c^2}{a^2} \geq \dfrac{2b}{a}$
$\dfrac{c^2}{a^2} + \dfrac{a^2}{b^2} \geq \dfrac{2c}{b}$
Cộng vế theo vế ta được:
$2\left(\dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\right) \geq 2\left(\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}\right)$
$\Leftrightarrow \dfrac{a^2}{b^2} + \dfrac{b^2}{c^2} +\dfrac{c^2}{a^2}\geq \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$