Bài 4.8 trang 104 SBT đại số 10

Đề bài

Cho a, b, c là những số dương. Chứng minh rằng 

\(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{{a + b + c}}\)

Lời giải chi tiết

\((a + b + c)(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}) \) \(= 3+ (\dfrac{a}{b} + \dfrac{b}{a}) + (\dfrac{a}{c} + \dfrac{c}{a}) + (\dfrac{b}{c} + \dfrac{c}{b})\)

Áp dụng BĐT Cô-si ta có:

\(\begin{array}{l}
\dfrac{a}{b} + \dfrac{b}{a} \ge 2\sqrt {\dfrac{a}{b}.\dfrac{b}{a}} = 2\\
\dfrac{a}{c} + \dfrac{c}{a} \ge 2\sqrt {\dfrac{a}{c}.\dfrac{c}{a}} = 2\\
\dfrac{b}{c} + \dfrac{c}{b} \ge 2\sqrt {\dfrac{b}{c}.\dfrac{c}{b}} = 2
\end{array}\)

\(\begin{array}{l}
\Rightarrow \left( {a + b + c} \right)\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
\ge 3 + 2 + 2 + 2 = 9\\
\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{{a + b + c}}
\end{array}\)

Cách khác:

\(\begin{array}{l}
a + b + c \ge 3\sqrt[3]{{abc}}\\
\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge 3\sqrt[3]{{\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}}} = \dfrac{3}{{\sqrt[3]{{abc}}}}\\
\Rightarrow \left( {a + b + c} \right)\left( {\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}} \right)\\
\ge 3\sqrt[3]{{abc}}.\dfrac{3}{{\sqrt[3]{{abc}}}} = 9\\
\Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} \ge \dfrac{9}{{a + b + c}}
\end{array}\)