チェビシェフの和の不等式
a 1 b 1 + ⋯ + a n b n ≥ a 1 b 3 + a 2 b 4 + ⋯ + a n b 2 {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{3}+a_{2}b_{4}+\cdots +a_{n}b_{2}\,}
⋮ {\displaystyle \vdots \,}
a 1 b 1 + ⋯ + a n b n ≥ a 1 b n + a 2 b 1 + ⋯ + a n b n − 1 {\displaystyle a_{1}b_{1}+\cdots +a_{n}b_{n}\geq a_{1}b_{n}+a_{2}b_{1}+\cdots +a_{n}b_{n-1}}
となる。両辺それぞれについて総和を取って、
n ( a 1 b 1 + ⋯ + a n b n ) ≥ ( a 1 + ⋯ + a n ) ( b 1 + ⋯ + b n ) ; {\displaystyle n(a_{1}b_{1}+\cdots +a_{n}b_{n})\geq (a_{1}+\cdots +a_{n})(b_{1}+\cdots +b_{n});}
これを n 2 {\displaystyle n^{2}} で割ると、以下の不等式が得られる。
( a 1 b 1 + ⋯ + a n b n ) n ≥ ( a 1 + ⋯ + a n ) n ⋅ ( b 1 + ⋯ + b n ) n . {\displaystyle {\frac {(a_{1}b_{1}+\cdots +a_{n}b_{n})}{n}}\geq {\frac {(a_{1}+\cdots +a_{n})}{n}}\cdot {\frac {(b_{1}+\cdots +b_{n})}{n}}.}