cho$x,y> 0,x+y=1.tìm GTNN của$
$A=\left ( x+\frac{1}{x} \right )^{2}+\left ( y+\frac{1}{y} \right )^{2}$
Áp dụng AM-GM ta có:
$(x+\frac{1}{x})^{2}+(y+\frac{1}{y})^{2}\leq \frac{\left [ (x+\frac{1}{x})+(y+\frac{1}{y}) \right ]^{2}}{2}$
=$\frac{\left [(x+y)+(\frac{x+y}{xy})\right]^{2}}{2}$
Mà $xy\leq \frac{(x+y)^{2}}{4}=\frac{1}{4}\Rightarrow \frac{1}{xy}\geq 4\Rightarrow \frac{x+y}{xy}\geq 4$(vì x+y=1)
Do đó A=$(x+\frac{1}{x})^{2}+(y+\frac{1}{y})^{2}\leq \frac{\left [ (x+\frac{1}{x})+(y+\frac{1}{y}) \right ]^{2}}{2}$=$\frac{(1+4)^{2}}{4}$=$\frac{25}{2}$
Vậy MinA=$\frac{25}{2}\Leftrightarrow x=y=\frac{1}{2}$