Question

CMR: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+c+b\) với mọi số dương a,b,c

Answer 1

\(\dfrac{bc}{a}+\dfrac{ac}{b}=\dfrac{b^2c+a^2c}{ab}=\dfrac{c\left(a^2+b^2\right)}{ab}\ge\dfrac{c.2ab}{ab}=2c\)

\(\dfrac{ac}{b}+\dfrac{ab}{c}=\dfrac{ac^2+ab^2}{bc}=\dfrac{a\left(b^2+c^2\right)}{bc}\ge\dfrac{a.2bc}{bc}=2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}=\dfrac{a^2b+c^2b}{ac}=\dfrac{b\left(a^2+b^2\right)}{ac}\ge\dfrac{b.2ac}{ac}=2b\)

Cộng vế theo vế:

\(2\left(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\right)\ge2\left(a+b+c\right)\)

\(\Rightarrow\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)