Question

Cho các số dương a, b, c thỏa mãn a + b + c = 1. Chứng mình rằng: \(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6\)

Answer 1

Áp dụng BĐT cosi:

\(VT=\left(\dfrac{a^2}{b}+\dfrac{1}{b}\right)+\left(\dfrac{b^2}{c}+\dfrac{1}{c}\right)+\left(\dfrac{c^2}{a}+\dfrac{1}{a}\right)\\ \Leftrightarrow VT\ge2\sqrt{\dfrac{a^2}{b^2}}+2\sqrt{\dfrac{b^2}{c^2}}+2\sqrt{\dfrac{c^2}{a^2}}\\ \Leftrightarrow VT\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\ge2\cdot3\sqrt[3]{\dfrac{abc}{abc}}=6\)

Dấu \("="\Leftrightarrow a=b=c=\dfrac{1}{3}\)