作业帮是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 10:15:11
![是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy](/uploads/image/z/5256010-10-0.jpg?t=%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E5%B8%B8%E6%95%B0c%2C%E4%BD%BF%E5%BE%97%E4%B8%8D%E7%AD%89%E5%BC%8Fx%2F%282x%2By%29%2By%2F%28x%2B2y%29%E2%89%A4c%E2%89%A4x%2F%28x%2B2y%29%2By%2F%282x%2By%29%E5%AF%B9%E4%BB%BB%E6%84%8F%E6%AD%A3%E6%95%B0x%2Cy%E6%81%92%E6%88%90%E7%AB%8B%5Bx%2F%282x%2By%29+%2By%2F%28x%2B2y%29%5D-%5Bx%2F%28x%2B2y%29+%2By%2F%282x%2By%29%5D+%3D%5Bx%28x%2B2y%29%2By%282x%2By%29-x%282x%2By%29-y%28x%2B2y%29%5D%2F%5B%28x%2B2y%29%282x%2By%29%5D+%3D%5Bx%5E2%2B2xy%2B2xy%2By%5E2-2x%5E2-xy-xy-2y%5E2%5D%2F%5B%28x%2B2y%29%282x%2By%29%5D+%3D%282xy)
是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy
是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立
[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)]
=[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)]
=[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)]
=(2xy-x^2-y^2)/[(x+2y)(2x+y)]
=-(x-y)^2/[(x+2y)(2x+y)]≤0{对任意正数x,y}
----------------------------------------------------------------
其中,x=y时,等号成立
x=y时,x/(2x+y) +y/(x+2y)=x/3x+y/3y=1/3+1/3=2/3
所以,c=2/3时,
不等式x/(2x+y) +y/(x+2y)
就是为什x=y的时候成立?x=y不就等于零了?
-(x-y)^2/[(x+2y)(2x+y)]≤0
(x-y)^2/[(x+2y)(2x+y)]>=0
那x=y不等号成立了只能说明
(x-y)^2/[(x+2y)(2x+y)]=0
但不是说(x-y)^2/[(x+2y)(2x+y)]要>=0吗?
是否存在常数c,使得不等式x/(2x+y)+y/(x+2y)≤c≤x/(x+2y)+y/(2x+y)对任意正数x,y恒成立[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)] =[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)] =[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)] =(2xy
[x/(2x+y) +y/(x+2y)]-[x/(x+2y) +y/(2x+y)]
=[x(x+2y)+y(2x+y)-x(2x+y)-y(x+2y)]/[(x+2y)(2x+y)]
=[x^2+2xy+2xy+y^2-2x^2-xy-xy-2y^2]/[(x+2y)(2x+y)]
=(2xy-x^2-y^2)/[(x+2y)(2x+y)]
=-(x-y)^2/[(x+2y)(2x+y)]≤0{对任意正数x,y}
这部分是说,只要x>0、y>0,右边的式子总是大于或等于左边的式子.
其中,x=y时,等号成立
x=y时,x/(2x+y) +y/(x+2y)=x/3x+y/3y=1/3+1/3=2/3
所以,c=2/3时,
不等式x/(2x+y) +y/(x+2y)
x=y时,等号成立是说不等号中的等号成立
即[x/(2x+y) +y/(x+2y)]=[x/(x+2y) +y/(2x+y)]
设2x+y=3m,2y+x=3n 且m,n均为正数
则x=2m-n,y=2n-m
所以x/(2x+y)+y/(x+2y)=(2m-n)/3m+(2n-m)/3n
=4/3-1/3(n/m+m/n)≤4/3-1/3*2=2/3
x/(x+2y)+y/(2x+y)=(2m-n)/3n+(2n-m)/3m=2/3(n/m+m/n)-2/3≥2/3*2-2/3=2/3
所以c=2/3