Ta có : \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-a-b-c\)
= \(\frac{ab-ac}{c}+\frac{bc-ab}{a}+\frac{ca-bc}{b}\)
= \(\frac{ab\left(ab-ac\right)}{abc}+\frac{\left(bc\left(bc-ab\right)\right)}{abc}+\frac{ca\left(ca-bc\right)}{abc}\)
= \(\frac{a^2b\left(b-c\right)+b^2c\left(c-a\right)+c^2a\left(a-b\right)}{abc}\) \(\ge0\)
Do a,b,c > 0
Cách 2 . Áp dụng bất đẳng thức Cauchy , ta có :
\(\frac{ab}{c}+\frac{bc}{a}\ge2.\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2b\)
\(\frac{bc}{a}+\frac{ca}{b}\ge2c\)
\(\frac{ca}{b}+\frac{ab}{c}\ge2a\)
Cộng vế theo vế => \(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)
=> \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)
Đẳng thức xảy ra <=> a = b = c
