已知函数f(x)=tanx,x∈(0,π2).若x1,x2∈(0,π2),且x1≠x2,证明1/2[f(x1)+f(x2)]>f(x1+x22)

问题描述:

已知函数f(x)=tanx,x∈(0,

π
2
).若x1,x2∈(0,
π
2
),且x1≠x2
证明
1
2
[f(x1)+f(x2)]>f(
x1+x2
2

证明:tanx1+tanx2=

sinx1
cosx1
+
sinx2
cosx2
=
sinx1cosx2+cosx1sinx2
cosx1cosx2

=
sin(x1+x2)
cosx1cosx2
=
2sin(x1+x2)
cos(x1+x2)+cos(x1-x2)

∵x1,x2∈(0,
π
2
),x1≠x2
∴2sin(x1+x2)>0,cosx1cosx2>0,且0<cos(x1-x2)<1,
从而有0<cos(x1+x2)+cos(x1-x2)<1+cos(x1+x2),
由此得tanx1+tanx2>=
2sin(x1+x2)
1+cos(x1+x2)
,∴
1
2
(tanx1+tanx2)>tg
x1+x2
2

1
2
[f(x1)+f(x2)]>f(
x1+x2
2
).