Đáp án:
$\begin{array}{l}
\dfrac{{{a^2} + {b^2}}}{2} - ab\\
= \dfrac{{{a^2} + {b^2} - 2ab}}{2}\\
= \dfrac{{{{\left( {a - b} \right)}^2}}}{2} \ge 0\\
\Rightarrow \dfrac{{{a^2} + {b^2}}}{2} \ge ab\\
Do:x \ge 0;y \ge 0\\
\Rightarrow \dfrac{{{{\left( {\sqrt x } \right)}^2} + {{\left( {\sqrt y } \right)}^2}}}{2} \ge \sqrt x .\sqrt y \\
\Rightarrow \dfrac{{x + y}}{2} \ge \sqrt {x.y}
\end{array}$
