17/08/01 15:08:14.92 MADJ3GR6.net
>>352
(xx+2yy+3) - 2(√2 -1)(x+2y+3) = (x+1-√2)^2 + 2(y+1-√2)^2 ≧ 0,
(xx+2yy+3) + 2(√2 +1)(x+2y+3) = (x+1+√2)^2 + 2(y+1+√2)^2 ≧ 0,
両辺を xx+2yy+3 >0 で割って
-(√2 -1)/2 ≦ (x+2y+3)/(xx+2yy+3) ≦ (√2 +1)/2,
でござるか。
17/08/01 15:08:14.92 MADJ3GR6.net
>>352
(xx+2yy+3) - 2(√2 -1)(x+2y+3) = (x+1-√2)^2 + 2(y+1-√2)^2 ≧ 0,
(xx+2yy+3) + 2(√2 +1)(x+2y+3) = (x+1+√2)^2 + 2(y+1+√2)^2 ≧ 0,
両辺を xx+2yy+3 >0 で割って
-(√2 -1)/2 ≦ (x+2y+3)/(xx+2yy+3) ≦ (√2 +1)/2,
でござるか。