$#Ben$
Đáp án `+` Giải thích các bước giải:
`=>` $K\ge \dfrac{19}{3}$
`----------------------`
Đặt $x=a+2, y=b+1, a,b\ge 0$
$\to K=(a+2)^2+(b+1)^2+\dfrac{1}{a+2+b+1}+\dfrac{1}{b+1}$
$\to K=a^2+4a+4+b^2+2b+1+\dfrac{1}{a+b+3}+\dfrac{1}{b+1}$
$\to K=a^2+b^2+4a+2b+5+\dfrac{1}{a+b+3}+\dfrac{1}{b+1}$
$\to K\ge 0+0+4a+2b+5+\dfrac{1}{a+b+3}+\dfrac{1}{b+1}$
$\to K\ge 4a+2b+5+\dfrac{1}{a+b+3}+\dfrac{1}{b+1}$
$\to K\ge \dfrac19(35a+8b+33)+(\dfrac{a+b+3}{9}+\dfrac{1}{a+b+3})+((b+1)+\dfrac{1}{b+1})$
$\to K\ge \dfrac19(35\cdot 0+8\cdot 0+33)+2\sqrt{\dfrac{a+b+3}{9}\cdot\dfrac{1}{a+b+3}}+2\sqrt{(b+1)\cdot\dfrac{1}{b+1}}$
$\to K\ge \dfrac{19}{3}$
Dấu = xảy ra khi $a=b=0\to x=2,y=1$
