Question

Chứng minh: \(\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge2\) với mọi a; b khác 0

Answer 1

Xét hiệu 

\(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\) = \(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\cdot\frac{a}{b}\cdot\frac{b}{a}=\left(\frac{a}{b}-\frac{b}{a}\right)^2\)  \(\ge0\)

=> \(\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\ge0\Leftrightarrow\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge2\)

Dấu ' = ' xảy ra khi a = b