Question

chứng minh bất đẳng thức sau

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

Answer 1

Áp dụng bất đẳng thức AM-GM:

\(\frac{a^2}{b^2}+\frac{b^2}{c^2}\ge2\sqrt{\frac{a^2b^2}{b^2c^2}}=2\sqrt{\frac{a^2}{c^2}}=2\left|\frac{a}{c}\right|\ge\frac{2a}{c}\)

Chứng minh tương tự: \(\hept{\begin{cases}\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{2b}{a}\\\frac{a^2}{b^2}+\frac{c^2}{a^2}\ge\frac{2c}{b}\end{cases}}\)

Cộng theo vế: \(2\left(\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\right)\ge2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\)

\(\Rightarrow\frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\ge\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\)

Dấu "=" khi \(a=b=c\)