相加平均と相乗平均の関係
\[\begin{eqnarray}\frac{a+b}{2}-\sqrt{ab}&=&\frac{1}{2}\{(\sqrt{a})^{2}-2\sqrt{a}\sqrt{b}+(\sqrt{b})^{2}\}\\ &=& \frac{1}{2}(\sqrt{a}-\sqrt{b})^{2} \geqq 0 \end{eqnarray} \]よって,\[\frac{a+b}{2} \geqq \sqrt{ab} \]また,等号が成り立つのは,\[\sqrt{a}-\sqrt{b}=0 \]すなわち,$a=b$ のときである▪️
Mathematics is the language with which God has written the universe.