cho a,b,c>0. tìm giá trị nhỏ nhất của S = a + b^2 + c^3 + 1/abc. 08/09/2021 Bởi Arya cho a,b,c>0. tìm giá trị nhỏ nhất của S = a + b^2 + c^3 + 1/abc.
Đáp án: $S\ge 17\sqrt[17]{\dfrac{1}{235092492288}}$ Giải thích các bước giải: Ta có: $S=a+b^2+c^3+\dfrac{1}{abc}$ $\to S=(\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6})+(\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{b^2}{3})+(\dfrac{c^3}{2}+\dfrac{c^3}{2})+(\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc})$ $\to S=\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{c^3}{2}+\dfrac{c^3}{2}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}$ $\to S\ge 17\sqrt[17]{\dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{b^2}{3}\cdot \dfrac{b^2}{3}\cdot \dfrac{b^2}{3}\cdot \dfrac{c^3}{2}\cdot \dfrac{c^3}{2}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}}$ $\to S\ge 17\sqrt[17]{\dfrac{1}{235092492288}}$ Dấu = xảy ra khi : $\dfrac{a}{6}=\dfrac{b^2}{3}=\dfrac{c^3}{2}=\dfrac{1}{6abc}$ $\to b=\sqrt{\dfrac{a}{2}}, c=\sqrt[3]{\dfrac{a}{3}},\dfrac{a}{6}=\dfrac{1}{6abc}$ $\to \dfrac{a}{6}=\dfrac{1}{6a\cdot \sqrt{\dfrac{a}{2}}\cdot \sqrt[3]{\dfrac{a}{3}}}$ $\to a=\sqrt[17]{72}$ $\to b=\sqrt{\dfrac12\sqrt[17]{72}}, c=\sqrt[3]{\dfrac{\sqrt[17]{72}}{3}}$ Bình luận
Đáp án: $S\ge 17\sqrt[17]{\dfrac{1}{235092492288}}$
Giải thích các bước giải:
Ta có:
$S=a+b^2+c^3+\dfrac{1}{abc}$
$\to S=(\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6})+(\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{b^2}{3})+(\dfrac{c^3}{2}+\dfrac{c^3}{2})+(\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc})$
$\to S=\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{a}{6}+\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{b^2}{3}+\dfrac{c^3}{2}+\dfrac{c^3}{2}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}+\dfrac{1}{6abc}$
$\to S\ge 17\sqrt[17]{\dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{a}{6}\cdot \dfrac{b^2}{3}\cdot \dfrac{b^2}{3}\cdot \dfrac{b^2}{3}\cdot \dfrac{c^3}{2}\cdot \dfrac{c^3}{2}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}\cdot \dfrac{1}{6abc}}$
$\to S\ge 17\sqrt[17]{\dfrac{1}{235092492288}}$
Dấu = xảy ra khi :
$\dfrac{a}{6}=\dfrac{b^2}{3}=\dfrac{c^3}{2}=\dfrac{1}{6abc}$
$\to b=\sqrt{\dfrac{a}{2}}, c=\sqrt[3]{\dfrac{a}{3}},\dfrac{a}{6}=\dfrac{1}{6abc}$
$\to \dfrac{a}{6}=\dfrac{1}{6a\cdot \sqrt{\dfrac{a}{2}}\cdot \sqrt[3]{\dfrac{a}{3}}}$
$\to a=\sqrt[17]{72}$
$\to b=\sqrt{\dfrac12\sqrt[17]{72}}, c=\sqrt[3]{\dfrac{\sqrt[17]{72}}{3}}$