Đặt [tex]p=x+y+z=\frac{3}{2},q=xy+yz+zx,r=xyz\Rightarrow S=p^3-3pq+3r+r^2=\frac{27}{8}+r^2+3r-\frac{9}{2}q[/tex]
Theo BĐT Schur ta có: [tex]r\geq \frac{p(4q-p^2)}{9}=\frac{16q-9}{6}\Rightarrow q\leq \frac{3}{8}r+\frac{9}{16}\Rightarrow S\geq \frac{27}{8}-\frac{9}{2}(\frac{3}{8}r+\frac{9}{16})+r^2+3r=r^2+\frac{21}{16}r+\frac{27}{32}\geq \frac{27}{32}(r\geq 0)[/tex]
Dấu "=" xảy ra khi [tex]\left\{\begin{matrix} r=\frac{p(4q-p^2)}{9}\\ p=\frac{3}{2}\\ r=0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} p=\frac{3}{2}\\ 4q-p^2=0\\ r=0 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=y=\frac{3}{4}\\ z=0 \end{matrix}\right.[/tex]
Vậy Min S = [tex]\frac{27}{32}[/tex]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
