作业帮已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1,X2,证明：（1）当a小于等于0时,{[f(X1)+f(X2)]/2}>f[(X1+X2)/2](2)当a小于等于4时,|f'(x1)-f'(x2)|>|x1-x2|
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 01:43:52
![已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1,X2,证明：（1）当a小于等于0时,{[f(X1)+f(X2)]/2}>f[(X1+X2)/2](2)当a小于等于4时,|f'(x1)-f'(x2)|>|x1-x2|](/uploads/image/z/11498516-44-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%87%BD%E6%95%B0f%28x%29%3Dx%E7%9A%84%E5%B9%B3%E6%96%B9%2B%EF%BC%882%2FX%EF%BC%89%2BalnX%28X%3E0%29%2Cf%28x%29%E5%AF%BC%E5%87%BD%E6%95%B0%E6%98%AFf%27%28x%29.%E5%AF%B9%E4%BB%BB%E6%84%8F%E4%B8%A4%E4%B8%AA%E4%B8%8D%E7%AD%89%E7%9A%84%E6%AD%A3%E6%95%B0X1%2CX2%2C%E8%AF%81%E6%98%8E%EF%BC%9A%EF%BC%881%EF%BC%89%E5%BD%93a%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8E0%E6%97%B6%2C%7B%5Bf%28X1%29%2Bf%28X2%29%5D%2F2%7D%3Ef%5B%28X1%2BX2%29%2F2%5D%282%29%E5%BD%93a%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8E4%E6%97%B6%2C%7Cf%27%28x1%29-f%27%28x2%29%7C%3E%7Cx1-x2%7C)
已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1,X2,证明：（1）当a小于等于0时,{[f(X1)+f(X2)]/2}>f[(X1+X2)/2](2)当a小于等于4时,|f'(x1)-f'(x2)|>|x1-x2|
已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1,X2,证明：
（1）当a小于等于0时,{[f(X1)+f(X2)]/2}>f[(X1+X2)/2]
(2)当a小于等于4时,|f'(x1)-f'(x2)|>|x1-x2|
已知函数f(x)=x的平方+（2/X）+alnX(X>0),f(x)导函数是f'(x).对任意两个不等的正数X1,X2,证明：（1）当a小于等于0时,{[f(X1)+f(X2)]/2}>f[(X1+X2)/2](2)当a小于等于4时,|f'(x1)-f'(x2)|>|x1-x2|
{[f(X1)+f(X2)]/2}-f[(X1+X2)/2]
=(x1^2+2/x1+alnx1+x2^2+2/x2+alnx2)/2-[(x1+x2)^2/4+2/(x1+x2)/2+aln(x1+x2)/2]
=[(x1^2+x2^2）/2-(x1+x2)^2/4)]+[（2/x1+2/x2)/2-2/(x1+x2)/2）]+[（alnx1+alnx2）/2-aln(x1+x2)/2]
[(x1^2+x2^2）/2-(x1+x2)^2/4)]=(x1-x2)^2/4≥0
（2/x1+2/x2)/2-2/(x1+x2)/2）=(x1-x2)^2/[x1x2(x1+x2)]≥0
（alnx1+alnx2）/2-aln(x1+x2)/2=aln[√(x1x2)/(x1+x2)/2]
当x1,x2>0
显然(x1+x2)/2≥2√(x1x2)/2=√(x1x2)
所以ln[√(x1x2)/(x1+x2)/2]<=ln1=0
当a<=0时,aln[√(x1x2)/(x1+x2)/2]≥0
所以
{[f(X1)+f(X2)]/2}-f[(X1+X2)/2]≥0
则{[f(X1)+f(X2)]/2}>f[(X1+X2)/2]
|f'(x1)-f'(x2)|=2x1-2/x1^2+a/x1-(2x2-2/x2^2+a/x2)|
=|2(x1-x2)-2(1/x1^2-1/x2^2)+a*(1/x1-1/x2)|
=|2+2(x1+x2)/x1^2x2^2-a/x1x2||x1-x2|
而
|2+2(x1+x2)/x1^2x2^2-a/x1x2|
=|2(x1^2x2^2+2(x1+x2)-ax1x2))/x1^2x2^2|
x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√(x1x2)-ax1x2
当0
4√(x1x2)>ax1x2
则x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√(x1x2)-ax1x2>=0
当x1x2>1时,
则x1^2x2^2+2(x1+x2)-ax1x2≥x1^2x2^2+4√(x1x2)-ax1x2
=(x1x2-a/2)^2+4√(x1x2)-a^2/4
x1x2>1,a≤4时,4√(x1x2)-a^2/4>0
所以,(x1x2-2)^2+4√(x1x2)-4>=0
则x1^2x2^2+4√(x1x2)-ax1x2>=0
则x1^2x2^2+2(x1+x2)-ax1x2≥0
即2(x1^2x2^2+2(x1+x2)-ax1x2))>=x1^2x2^2
综合得：|f'(x1)-f'(x2)|>|x1-x2|
我给你回答第一问：
f'(x).=2x-1/x^2+a/x
f"(x)=2+2/x^3-a/x^2
=2+1/x^2*(2/x-a)
当a小于等于零时，f"(x)恒大于等于零；所以f(x)为凹函数，所以得证！