절대부등식 보기

ㄱ. 산술 기하평균에서 $\displaystyle a + \frac{1}{a} \geq 2 \sqrt{a \times {\frac{1}{a}}} = 2 \therefore a \geq 2$ ㄴ. $a ^{2} + b ^{2} + c ^{2} - ( a b + bc + ca )$ $\displaystyle = \frac{1}{2} \left\{ ( a - b ) ^{2} + ( b - c ) ^{2} + ( c - a ) ^{2} \right\} > 0$ $\therefore a ^{2} + b ^{2} + c ^{2} > a b + bc + ca$ $\because a , b , c$가 서로 다른 양수 ㄷ. $\left| a + b \right| ^{2} - ( \left| a \right| + \left| b \right| ) ^{2} = 2 ( a b - \left| a b \right| ) = 2 ( a b - a b ) = 0$ $\therefore \left| a + b \right| = \left| a \right| + \left| b \right|$