... x1)+ (y 2 − x 2 ) 2 + (ey 2 /2 − ex 2 /2 )(y 2 − x 2 )≥ 0,vàF 2 (y) − F 2 (x), y − x = (y1− x1) 2 + 3(y 2 − x 2 )(y1− x1)+ 3(y 2 − x 2 ) 2 + (ey 2 − ex 2 /2 )(y 2 − x 2 )≥ 0.Do ... R 2 → R 2 2 được cho bởiF (x) =x1+ x 2 x 2 + ex 2 /2 − 2 x1+ 2x 2 x1+ 3x 2 + ex 2 − 1.Khi đó với mọi x, y ∈ Rkta cóF1(y) − F1(x), y − x = (y1− x1) 2 + (y 2 − x 2 )(y1− ... (−0.4373374051 023 18,−0.808736590634006)T0.00000011 92 17796803 42 25 (−0.437337404356388,−0.80873 629 3018647)T0.0000000596356 025 3 327 26 (−0.437337403855591,−0.808736144 322 769)T0.000000 029 87 128 665016 27 (−0.437337403701 826 ,−0.808736069890305)T0.00000001491 420 5655551...