Đáp án:
`\text{Min}` `m(m/2+\frac{1}{np})+n(n/2+\frac{1}{mp})+p(p/2+\frac{1}{mn})=9/2`
Giải thích các bước giải:
Ta có:
`\sum [m(m/2+\frac{1}{np})]`
`=\sum (\frac{m^2}{2}+\frac{m}{np})`
`=\frac{\sum m^2}{2}+\sum \frac{m}{np}`
`>=\frac{3\root{3}{(xyz)^2}}{2}+\frac{3}{\root{3}{xyz}}`
Lại có:
`\frac{3\root{3}{(xyz)^2}}{2}+\frac{3}{\root{3}{xyz}}-9/2`
`=\frac{2(\root{3}{xyz}-1)^2(\root{3}{xyz}^2+2)}{2\root{3}{xyz}}>=0`
`=>\frac{3\root{3}{(xyz)^2}}{2}+\frac{3}{\root{3}{xyz}}>=9/2`
`=>m(m/2+\frac{1}{np})+n(n/2+\frac{1}{mp})+p(p/2+\frac{1}{mn})>=9/2`
Dấu bằng xảy ra khi `m=n=p=1`
