一道高二不等式证明

证：
a>1,b>1，设m=a-1>0,n=b-1>0，则
a=1+m,b=1+n
(m-n)^2≥0，2(m-n)^2≥0,m^2+n^2≥2mn

m*(n-1)^2≥0, n*(m-1)^2≥0
mn^2+m≥2mn......(1)
nm^2+n≥2mn......(2)
(1)+(2),得
(mn^2+m)+(nm^2+n)≥4mn
mn^2+nm^2+m+n≥4mn
(m+n)*mn+(m+n)≥4mn
(m+n)*(2mn-mn)+(m+n)≥4mn
(m+n)*(m^2+n^2-mn)+(m+n)≥4mn
m^3+n^3+m+n≥4mn
m^3+n^3+m+n+2(m-n)^2≥4mn
m^3+n^3+m+n+2m^2+2n^2-4mn≥4mn
m^3+n^3+m+n+2m^2+2n^2≥8mn
（m^3+2m^2+m）+（n^3+2n^2+n）≥8mn
m(1+m)^2+n(1+n)^2≥8mn
(1+m)^2/n+(1+n)^2/m≥8
a^2/(b-1)+b^2/(a-1)≥8

证明：设m=a-1,n=b-1(m,n>0),则a=m+1,b=n+1
求证(m+1)^2/n+(n+1)^2/m>=8
因为m+1>=2根号m
所以(m+1)^2>=4m
(m+1)^2/n>=4m/n
同理(n+1)^2/m>=4n/m
所以(m+1)^2/n+(n+1)^2/m>=4(m/n+n/m)>=4*2根号(m/n*n/m)=8
证毕！