Giải thích các bước giải:
ĐK: $x,y>0$
Ta có:
$\begin{array}{l}
x + y\\
= \dfrac{x}{2} + \dfrac{x}{2} + \dfrac{y}{3} + \dfrac{y}{3} + \dfrac{y}{3}\\
\ge 5\sqrt[5]{{\dfrac{x}{2}.\dfrac{x}{2}.\dfrac{y}{3}.\dfrac{y}{3}.\dfrac{y}{3}}}\left( {BDT - Cauchy} \right)\\
= 5\sqrt[5]{{\dfrac{{{x^2}{y^3}}}{{{2^2}{{.3}^3}}}}}\\
= 5\sqrt[5]{{\dfrac{{108}}{{108}}}}\\
= 5
\end{array}$
Dấu bằng xảy ra:
$\begin{array}{l}
\left\{ \begin{array}{l}
\dfrac{x}{2} = \dfrac{y}{3}\\
{x^2}{y^3} = 108
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
{\left( {\dfrac{2}{3}y} \right)^2}{y^3} = 108
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
{y^5} = 243
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{2}{3}y\\
y = 3
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 3
\end{array} \right.
\end{array}$
Vậy ta có đpcm
