Question

cho ba số a,b,c >0. cmr: a/b+ b/c+ c/a>=a +b + c

Answer 1

\(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge a+b+c\)

\(\Leftrightarrow\dfrac{a^2c+b^2a+c^2b}{abc}\ge a+b+c\)

\(\Leftrightarrow a^2c+b^2c+c^2b\ge abc\left(a+b+c\right)\)

\(\Leftrightarrow a^2c+b^2c+c^2b\ge a^2bc+ab^2c+abc^2\)

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

\(\Leftrightarrow a^2c\left(1-b\right)+b^2c\left(1-a\right)+c^2b\left(1-c\right)\ge0\)

-Sửa đề: \(0< a,b,c\le1\) thì BĐT mới đúng.