求数学题解1设x>0y>0,证明
因为x>0y>0
所以[(x2+y2)^1/2]^6=[x^2+y^2]^3=x^6+y^6+3*x^2*y^4+3*x^4*y^2
[(x3+y3)^1/3]^6=[x^3+y^3]^2=x^6+y^6+2*x^3*y^3
[(x2+y2)^1/2]^6-[(x3+y3)^1/3]^6
=[x^6+y^6+3*x^2*y^4+3*x^4*y^2]-[x^6+y^6+2*x^3*y^3]
=3*x^4*y^2+3*x^2*y^4-2*x^3*y^3
=(x^2*y^2){(x-y)^2+2*x^2+2*y^2]
因为x>0 y>0
所以(x^2*y^2){(x-y)^2+2*x^2+2*y...全部
因为x>0y>0
所以[(x2+y2)^1/2]^6=[x^2+y^2]^3=x^6+y^6+3*x^2*y^4+3*x^4*y^2
[(x3+y3)^1/3]^6=[x^3+y^3]^2=x^6+y^6+2*x^3*y^3
[(x2+y2)^1/2]^6-[(x3+y3)^1/3]^6
=[x^6+y^6+3*x^2*y^4+3*x^4*y^2]-[x^6+y^6+2*x^3*y^3]
=3*x^4*y^2+3*x^2*y^4-2*x^3*y^3
=(x^2*y^2){(x-y)^2+2*x^2+2*y^2]
因为x>0 y>0
所以(x^2*y^2){(x-y)^2+2*x^2+2*y^2]>0
得到[(x2+y2)^1/2]^6-[(x3+y3)^1/3]^6>0
得到[(x2+y2)^1/2]^6>[(x3+y3)^1/3]^6
得到(x2+y2)^1/2>(x3+y3)^1/3
且全部证明 必须满足x>0 y>0。
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