a) Áp dụng BĐT Cô si cho 2 số dương ta có :
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}\ge2b\)
b) \(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)
CMTT như câu a ta đc :
\(\frac{ab}{c}+\frac{bc}{a}\ge2b;\frac{ab}{c}+\frac{ca}{b}\ge2a;\frac{bc}{a}+\frac{ac}{b}\ge2c\)
Do đó : \(\frac{ab}{c}+\frac{bc}{a}+\frac{ab}{c}+\frac{ca}{b}+\frac{bc}{a}+\frac{ca}{b}\ge2a+2b+2c\)
\(\Rightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\left(đpcm\right)\)
a. Áp dung BĐT AM-GM:
\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{ab}{c}.\frac{bc}{a}}=2\sqrt{b^2}=2b\)
b. Áp dung BĐT AM-GM:
\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)
\(\frac{bc}{a}+\frac{ca}{b}\ge2c\)
\(\frac{ca}{b}+\frac{ab}{c}\ge2a\)
\(\Rightarrow2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)
\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\)
Xảy ra đẳng thức khi \(a=b=c>0\)
