VT = \(\left(\frac{a^2}{b^2}-2.\frac{a}{b}+1\right)+\left(\frac{b^2}{a^2}-2.\frac{b}{a}+1\right)+2-\frac{a}{b}-\frac{b}{a}\)
= \(\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}-1\right)^2+\left(2-\frac{a}{b}-\frac{b}{a}\right)\)
Nhận xét: \(2-\frac{a}{b}-\frac{b}{a}=1+\frac{a}{b}.\frac{b}{a}-\frac{a}{b}-\frac{b}{a}=\left(\frac{a}{b}.\frac{b}{a}-\frac{a}{b}\right)+\left(1-\frac{b}{a}\right)\)
= \(\frac{a}{b}.\left(\frac{b}{a}-1\right)+\left(1-\frac{b}{a}\right)=\left(1-\frac{b}{a}\right).\left(1-\frac{a}{b}\right)=\left(\frac{a}{b}-1\right).\left(\frac{b}{a}-1\right)\)
=> VT = \(\left(\frac{a}{b}-1\right)^2+\left(\frac{b}{a}-1\right)^2+\left(\frac{a}{b}-1\right)\left(\frac{b}{a}-1\right)\)
= \(\left(\frac{a}{b}-1\right)^2+2.\left(\frac{a}{b}-1\right).\frac{\left(\frac{b}{a}-1\right)}{2}+\frac{\left(\frac{b}{a}-1\right)^2}{4}+\frac{3.\left(\frac{b}{a}-1\right)^2}{4}\)
= \(\left(\left(\frac{a}{b}-1\right)+\frac{\left(\frac{b}{a}-1\right)}{2}\right)^2+\frac{3.\left(\frac{b}{a}-1\right)^2}{4}\ge\frac{3.\left(\frac{b}{a}-1\right)^2}{4}\ge0\) với mọi a; b
=> đpcm
Dấu = khi \(\frac{a}{b}-1=-\frac{\frac{b}{a}-1}{2}\) và \(\frac{b}{a}-1=0\) <=> a = b
