Question

Cho a,b,c >0
C/m:  a,\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)
         b, \(\frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b}\ge a+b+c\)

Answer 1

a/ \(\frac{ab}{c}+\frac{bc}{a}\ge2b\)

\(\Leftrightarrow a^2b+bc^2\ge2abc\)

\(\Leftrightarrow a^2b+bc^2-2abc\ge0\)

\(\Leftrightarrow\left(a\sqrt{b}-c\sqrt{b}\right)^2\ge0\)(đúng)

\(\RightarrowĐPCM\)

Answer 2

b/ Áp dụng câu a ta có

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

\(\frac{ab}{c}+\frac{ca}{b}\ge2a\)

\(\frac{bc}{a}+\frac{ac}{b}\ge2c\)

Cộng 3 cái đó vế theo vế được

Answer 3

\(2\left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\ge2\left(a+b+c\right)\)

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