Question

Cho a,b,c >0. Chứng minh rằng: \(\dfrac{bc}{a}+\dfrac{ac}{b}+\dfrac{ab}{c}\ge a+b+c\)

Answer 1

áp dụng BĐT cô si cho 2 số ta có

\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{bc}{a}.\dfrac{ac}{b}}\)

⇔\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{c^2}=2c\)

TT ta có \(\dfrac{ac}{b}+\dfrac{ab}{c}\ge2a\)

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2b\)

cộng từng vế 3 BĐT trên

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

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