Question

Cho a,b,c > 0

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

Answer 1

Ta có ; \(\frac{a^2}{b^2}+\frac{b^2}{a}-\left(\frac{a}{b}+\frac{b}{a}\right)\ge\Leftrightarrow\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-2\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)\ge0\)

Lại có : \(\frac{a}{b}+\frac{b}{a}\ge2\)

Suy ra \(\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)-2\left(\frac{a}{b}+\frac{b}{a}\right)+2\ge0\)

\(\Leftrightarrow\left(\frac{a^2}{b^2}-\frac{2a}{b}+1\right)+\left(\frac{b^2}{a^2}-\frac{2b}{a}+1\right)\ge0\)

\(\Leftrightarrow\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}-1\right)^2\ge0\) (luôn đúng)

Vậy bđt ban đầu dc chứng minh