證:(1) ∵ tÎR, t ¹ –1,
∴ ⊿ = (–c
2a)
2 – 16c
2 = c
4a
2 – 16c
2³ 0 ,
∵ c ¹ 0, ∴c
2a
2³ 16 , ∴| ac | ³ 4.
(2) 由 f ( x ) =" 1" –
,
法1. 設(shè)–1 < x
1 < x
2, 則f (x
2) – f ( x
1) =" 1–"
–1 +
=
.
∵ –1 < x
1 < x
2, ∴ x
1 – x
2 < 0, x
1 + 1 > 0, x
2 + 1 > 0 ,
∴f (x
2) – f ( x
1) < 0 , 即f (x
2) < f ( x
1) , ∴x ³ 0時,f ( x )單調(diào)遞增.
法2. 由f ` ( x ) =
> 0 得x ¹ –1, ∴x > –1時,f ( x )單調(diào)遞增.
(3)∵f ( x )在x > –1時單調(diào)遞增,| c | ³
> 0 ,
∴f (| c | ) ³ f (
) =
=
f ( | a | ) + f ( | c | ) =
+
>
+
=1.
即f ( | a | ) + f ( | c | ) > 1.