Godzio:
| 2 | | x2+1 | |
| + |
| > 2 /*2(x2+1) |
| x2+1 | | 2 | |
4 + (x
2+1)
2 > 4(x
2+1)
x
4 +2x
2 + 1 − 4x
2 > 0
−3x
4 + 2x
2 + 1 > 0
i tu t = x
2 i −3t
2 + 2t + 1 > 0 lub jak wolisz to:
−3x
4 + 3x
2 − x
2 + 1 > 0
−3x
2(x
2−1) −(x
2−1)>0
−(x
2−1)(x
2+1) >0
(x−1)(x+1)(x
2+1)<0 x
2 + 1 > 0 dla x∊R
(x−1)(x+1) < 0
x∊(−1,1)