相加平均・相乗平均の不等式

arithmetic and geometric means; AM-GM inequality

$ \frac{a+b}{2} \ge \sqrt{ab} \quad(a,b > 0)

code:tex

\begin{aligned}

(\sqrt{a} - \sqrt{b})^2 &= a - 2\sqrt{ab} + b \ge 0\\

&\iff a + b \ge 2\sqrt{ab}\\

&\iff \frac{a+b}{2} \ge \sqrt{ab}

\end{aligned}

$ ab \le \left(\frac{a+b}{2}\right)^2

4変数に拡張する

code:tex

\begin{aligned}

ab\cdot cd &\le \left(\frac{a+b}{2}\right)^2\left(\frac{c+d}{2}\right)^2

= \left(\frac{a+b+c+d}{4}\right)^4

\end{aligned}

実際

code:tex

\frac{1}{3}(a+b+c)abc = \left(\frac{1}{3}(a+b+c)\right)^4\\

\iff abc = \left(\frac{a+b+c}{3}\right)^3