>>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,