Cho x>0,y>0 và x+y=6. Tìm GTNN của tổng $\frac{1}{x}$ + $\frac{1}{y}$ 25/08/2021 Bởi Camila Cho x>0,y>0 và x+y=6. Tìm GTNN của tổng $\frac{1}{x}$ + $\frac{1}{y}$
$x+y=6\\ \Rightarrow x=6-y\\ \dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{1}{6-y}+\dfrac{1}{y}\\ =\dfrac{y+6-y}{y(6-y)}\\ =\dfrac{6}{-y^2+6y}\\ =\dfrac{6}{-y^2+6y-9+9}\\ =\dfrac{6}{-(y-3)^2+9}$ Do $x>0;y>0\Rightarrow \dfrac{1}{x}+\dfrac{1}{y} \Rightarrow \dfrac{6}{-(y-3)^2+9}>0 \Rightarrow -(y-3)^2+9>0$ $\Rightarrow 0<-(y-3)^2+9\le 9\\ \Rightarrow \dfrac{6}{-(y-3)^2+9} \ge \dfrac{6}{9}=\dfrac{2}{3}$ Dấu “=” xảy ra $\Leftrightarrow \left\{\begin{array}{l} x=6-y\\ y-3=0\end{array} \right.\Leftrightarrow x=y=3$ Bình luận
$x+y=6\\ \Rightarrow x=6-y\\ \dfrac{1}{x}+\dfrac{1}{y}\\ =\dfrac{1}{6-y}+\dfrac{1}{y}\\ =\dfrac{y+6-y}{y(6-y)}\\ =\dfrac{6}{-y^2+6y}\\ =\dfrac{6}{-y^2+6y-9+9}\\ =\dfrac{6}{-(y-3)^2+9}$
Do $x>0;y>0\Rightarrow \dfrac{1}{x}+\dfrac{1}{y} \Rightarrow \dfrac{6}{-(y-3)^2+9}>0 \Rightarrow -(y-3)^2+9>0$
$\Rightarrow 0<-(y-3)^2+9\le 9\\ \Rightarrow \dfrac{6}{-(y-3)^2+9} \ge \dfrac{6}{9}=\dfrac{2}{3}$
Dấu “=” xảy ra $\Leftrightarrow \left\{\begin{array}{l} x=6-y\\ y-3=0\end{array} \right.\Leftrightarrow x=y=3$