Question

Cho a,b,c > 0. Chứng minh: a) \(\frac{ab}{c}\) +\(\frac{bc}{a}\) \(\ge2b\)

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

Answer 1

Câu b nhá mn

Answer 2

quá dễ BĐT_AM-GM sẽ cân tất cả

Answer 3

\(\frac{ab}{c}+\frac{bc}{a}\ge2\sqrt{\frac{abbc}{ac}}=2\sqrt{b^2}=2b;tươngtự:\left\{{}\begin{matrix}\frac{bc}{a}+\frac{ac}{b}\ge2\sqrt{\frac{abc^2}{ab}}=2c\\\frac{ac}{b}+\frac{ab}{c}\ge2\sqrt{\frac{a^2bc}{bc}}=2a.Cộngvếtheovếtađược:2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\right)\ge2\left(a+b+c\right)\end{matrix}\right.\)

\(\Leftrightarrow\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge a+b+c\left(\text{đpcm}\right)\)