Ta có: a, b, c>0⇒ \(\dfrac{\text{bc}}{\text{a}}\), \(\dfrac{\text{ac}}{\text{b}} , \dfrac{\text{ab}}{\text{c}}\)>0
Áp dụng bất đẳng thức Cô-si, ta có:
\(\dfrac{\text{bc}}{\text{a}} + \dfrac{\text{ac}}{\text{b}}\)≥2\(\sqrt{\dfrac{bc.ac}{a.b}}\) =2\(\sqrt{c^2}\) =2c (1)
\(\dfrac{ac}{b}+\dfrac{ab}{c}\) ≥2\(\sqrt{\dfrac{ac.ab}{b.c}}\) =2\(\sqrt{a^2}\)=2a (2)
\(\dfrac{ab}{c} + \dfrac{bc}{a}\) ≥2\(\sqrt{\dfrac{ab.bc}{c.a}}\) =2\(\sqrt{b^2}\) =2b (3)
Cộng theo vế (1), (2), (3), ta đc:
2(\(\dfrac{bc}{a} + \dfrac{ac}{b} + \dfrac{ab}{c})\) ≥2(a+b+c)
⇔\(\dfrac{bc}{a} + \dfrac{ac}{b} + \dfrac{ab}{c}\) ≥a+b+c (đpcm)
