Áp dụng BĐT $AM-GM$ ta có
$\frac{x}{1+x}=1-\frac{1}{1+x}\geq \frac{1}{y+2}+\frac{1}{z+3}\geq \frac{2}{\sqrt{(y+2)(z+3)}}$
$\frac{y+1}{y+2}=1-\frac{1}{y+2}\geq \frac{1}{x+1}+\frac{1}{z+3}\geq \frac{2}{\sqrt{(x+1)(z+3)}}$
$\frac{z+2}{z+3}=1-\frac{1}{z+3}\geq \frac{1}{x+1}+\frac{1}{y+2}\geq \frac{2}{\sqrt{(x+1)(y+2)}}$
Nhân theo vế các BĐT trên ta được
$\frac{x(y+1)(z+2)}{(x+1)(y+2)(z+3)}\geq \frac{8}{(x+1)(y+2)(z+3)}\Rightarrow x(y+1)(z+2)\geq 8$
Theo BĐT $AM-GM$ thì $8\leq x(y+1)(z+2)\leq \frac{(x+y+z+3)^3}{27}\Rightarrow x+y+z\geq 3$
Đặt $x+y+z=t$ $(t\geq 3)$
$\Rightarrow P=t+\frac{1}{t}=t+\frac{9}{t}-\frac{8}{t}\geq 2\sqrt{t.\frac{9}{t}}-\frac{8}{3}=\frac{10}{3}$
Vậy $P$ min $=\frac{10}{3}$. Dấu "=" $\Leftrightarrow x=2;y=1;z=0$
Mình nghĩ làm thế này dễ hiểu hơn:
Ta có: $1\geq \frac{1}{x+1}+\frac{1}{y+2}+\frac{1}{z+3}\geq \frac{(1+1+1)^{2}}{x+y+z+6}=\frac{9}{x+y+z+6}$
$\Rightarrow x+y+z+6\geq 9\Rightarrow x+y+x\geq 3$
Ta lại có: $x+y+z+\frac{1}{x+y+z}=\frac{8(x+y+z)}{9}+\frac{x+y+z}{9}+\frac{1}{x+y+z}\geq \frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}$
Vậy Pmin = $\frac{10}{3}$
- nguyenhongsonk612 yêu thích