Question

 \(x\ge2,\)\(x+y\ge3\). TÌM GTNN CỦA \(P=X^2+Y^2+\frac{1}{X}+\frac{1}{X+Y}\)

Answer 1

Lời giải:

Áp dụng BĐT AM-GM:

$x^2+4\geq 4x; y^2+1\geq 2y$

$\Rightarrow P=x^2+y^2+\frac{1}{x}+\frac{1}{x+y}$

$\geq 4x+2y+\frac{1}{x}+\frac{1}{x+y}-5$

$=[\frac{x+y}{9}+\frac{1}{x+y}]+[\frac{x}{4}+\frac{1}{x}]+\frac{131}{36}x+\frac{17}{9}y-5$

$\geq 2\sqrt{\frac{1}{9}}+2\sqrt{\frac{1}{4}}+\frac{17}{9}(x+y)+\frac{7}{4}x-5$

$\geq \frac{2}{3}+1+\frac{17}{9}.3+\frac{7}{4}.2-5=\frac{35}{6}$

Vậy $P_{\min}=\frac{35}{6}$ khi $x=2; y=1$