Đáp án:
`B_{min}={27}/4` khi `a=b=c=1/2`
Giải thích các bước giải:
Điều kiện `a;b;c>0`
Ta có:
`B=a^2+b^2+c^2 +1/a+1/b+1/c`
`=a^2+b^2+c^2+1/{8a}+1/{8a}+1/{8b}+1/{8b}+1/{8c}+1/{8c}+3/4 .(1/a+1/b+1/c)`
Áp dụng BĐT Cosi với các số dương ta có:
$a^2+\dfrac{1}{8a}+\dfrac{1}{8a}\ge 3\sqrt[3]{a^2 .\dfrac{1}{8a}.\dfrac{1}{8a}}=\dfrac{3}{4}$
Tương tự:
`b^2+1/{8b}+1/{8b}\ge 3/4`
`c^2+1/{8c}+1/{8c}\ge 3/4`
$\\$
$a+b+c\ge 3\sqrt[3]{abc}$
$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge 3\sqrt[3]{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}=3\sqrt[3]{\dfrac{1}{abc}}$
`=>(a+b+c)(1/a+1/b+1/c)\ge `$9\sqrt[3]{abc. \dfrac{1}{abc}}=9$
`=>1/a+1/b+1/c\ge 9/{a+b+c}`
Vì `a+b+c\le 3/2=>1/{a+b+c}\ge 2/3`
`=>1/a+1/b+1/c\ge 9. 2/3=6`
$\\$
`=>(a^2+1/{8a}+1/{8a})+(b^2+1/{8b}+1/{8b})+(c^2+1/{8c}+1/{8c})+3/4 .(1/a+1/b+1/c)`
`\ge 3/4+3/4+3/4+3/4. 6={27}/4`
`=>B\ge {27}/4`
Dấu "=" xảy ra khi:
$\quad \begin{cases}a^2=\dfrac{1}{8a}\\b^2=\dfrac{1}{8b}\\c^2=\dfrac{1}{8c}\\a+b+c=\dfrac{3}{2}\end{cases}$`=>`$\begin{cases}a^3=\dfrac{1}{8}\\b^3=\dfrac{1}{8}\\c^3=\dfrac{1}{8}\\a+b+c=\dfrac{3}{2}\end{cases}$
`=>a=b=c=1/2` (thỏa mãn)
Vậy $GTNN$ của $B$ bằng `{27}/4` khi `a=b=c=1/2`
