已知f(x)=√x,x∈(0,+∞),对于任意的x1,x2∈(0,+∞),求证[f(x1)+f(x2)]/2<f[(x1+x2)/2]成立。

利用
a的平方+b的平方>=2ab
{[f(x1)+f(x2)]/2}的平方
=(x1+x2+2√x1x2)/4
<=(x1+x2+ x1+x2)/4
=(x1+x2)/2
={ f[(x1+x2)/2]}的平方
即可证明结论成立

[f(x1)+f(x2)]/2= (√x1+√x2)/2
f[(x1+x2)/2]=[√(x1+x2)/2]
x1,x2∈(0,+∞),
(√x1^2+√x2^2-2√x1x2)>0

[(√x1+√x2)/2]^2<2(√x1^2+√x2^2)=2(x1+x2)

(√x1+√x2)/2<√(x1+x2)/2
[f(x1)+f(x2)]/2<f[(x1+x2)/2]

(√x1+√x2)^2>0 =>
x1+x2>2√x1x2 =>
x1+x2>√x1x2+(x1+x2)/2 =>
√(x1+x2)>(√x1+√x2)/√2 =>
√[(x1+x2)/2]>(√x1+√x2)/2 =>
f[(x1+x2)/2]>[f(x1)+f(x2)]/2