- 427 名前：１３２人目の素数さん mailto:sage [2009/08/25(火) 18:56:29 ]
- >>423
軸を45ﾟ回す。
x^2 + (1-y)^2 = (1/2)(x+y-1)^2 + (1/2)(x-y+1)^2 = u^2 + {v + (1/√2)}^2 ≧ {v + (1/√2)}^2,
(1-x)^2 + y^2 = (1/2)(x+y-1)^2 + (1/2)(x-y-1)^2 = u^2 + {v - (1/√2)}^2 ≧ {v - (1/√2)}^2,
よって
√[x^2 + (1-y)^2] ≧ |v + (1/√2)|,
√[(1-x)^2 + y^2] ≧ |v - (1/√2)|,
辺々たす。
(与式) ≧ |(1/√2) - (-1/√2)| = √2,
