Question

chứng minh bất đẳng thức

\(\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}+4\ge3\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\)

Answer 1

Đặt $\frac{a}{b}+\frac{b}{a}=t$

=>$t^2-3t+2$\geq$0$<=>$(t-1)(t-2)$\geq$0$<=>$(\frac{a}{b}+\frac{b}{a}-1)(\frac{a}{b}+\frac{b}{a}-2)$

<=>$\frac{(x-y)^2}{xy}\frac{(x-\frac{y}{2})^2+\frac{3y^2}{4}}{xy}\geq 0$

<=>$\frac{(x-y)^2.\left [ (x-\frac{y}{2})^2+\frac{3y^2}{4} \right ]}{x^2y^2}\geq 0$(hcđ)

=>đpcm