Ta có: [tex]P=\frac{x^2}{y}+\frac{y^2}{x}+\frac{1}{x+y}\geq \frac{(x+y)^2}{x+y}+\frac{1}{x+y}=x+y+\frac{1}{x+y}[/tex]
a) [tex]P\geq x+y+\frac{1}{x+y}\geq 2\sqrt{(x+y).\frac{1}{x+y}}=2[/tex]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x=y\\ x+y=1 \end{matrix}\right.\Leftrightarrow x=y=\frac{1}{2}[/tex]
b) Với x,y nguyên dương thì [tex]x,y\geq 1\Rightarrow x+y\geq 2\Rightarrow P\geq x+y+\frac{1}{x+y}=\frac{x+y}{4}+\frac{1}{x+y}+\frac{3}{4}(x+y)\geq 2\sqrt{\frac{x+y}{4}.\frac{1}{x+y}}+\frac{3}{4}.2=1+\frac{3}{2}=\frac{5}{2}[/tex]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} x+y=2\\ x=y \end{matrix}\right.\Leftrightarrow x=y=1[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
