Đáp án:
$min_S=15\dfrac{3^\tfrac{3}{5}}{2^\tfrac{4}{5}}\Leftrightarrow a=\dfrac{3^\tfrac{4}{5}}{2^\tfrac{2}{5}}$
Giải thích các bước giải:
$S=a^2+\dfrac{18}{\sqrt{a}}(a>0)\\ =a^2+\dfrac{18}{4\sqrt{a}}+\dfrac{18}{4\sqrt{a}}+\dfrac{18}{4\sqrt{a}}+\dfrac{18}{4\sqrt{a}}\\ \ge 5\sqrt[5]{a^2.\dfrac{18}{4\sqrt{a}}.\dfrac{18}{4\sqrt{a}}.\dfrac{18}{4\sqrt{a}}.\dfrac{18}{4\sqrt{a}}}(Cauchy)\\ =5\sqrt[5]{\dfrac{18^4}{4^4}}\\ =5\sqrt[5]{\dfrac{9^4}{2^4}}\\ =5\sqrt[5]{\dfrac{3^8}{2^4}}\\ =5\dfrac{3^\tfrac{8}{5}}{2^\tfrac{4}{5}}\\ =15\dfrac{3^\tfrac{3}{5}}{2^\tfrac{4}{5}}$
Dấu "=" xảy ra $\Leftrightarrow a^2=\dfrac{18}{4\sqrt{a}}$
$\Leftrightarrow a^\tfrac{5}{2}=\dfrac{9}{2}\\ \Leftrightarrow a=\sqrt[\tfrac{5}{2}]{\dfrac{9}{2}}=\dfrac{9^\tfrac{2}{5}}{2^\tfrac{2}{5}}=\dfrac{3^\tfrac{4}{5}}{2^\tfrac{2}{5}}$
