Question

不等式问题

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Answer 1

ab=1/c,设a+b=u>=2/√c,则
a^2+a+1+b^2+b+1=u^2+u+2-2/c,
(a^2+a+1)(b^2+b+1)
=1/c^2+a^2(b+1)+b^2(a+1)+(a+1)(b+1)
=1/c^2+u/c+u^2-2/c+1/c+u+1
=u^2+u+1+(u-1)/c+1/c^2,
不等式变为(u^2+u+2-2/c)/[u^2+u+1+(u-1)/c+1/c^2]>=(c^2+c)/(c^2+c+1),
又变为（u^2+u+2-2/c）（c^2+c+1)-[u^2+u+1+(u-1)/c+1/c^2](c^2+c)
=(c^2+c)[1-(u+1)/c-1/c^2]+u^2+u+2-2/c
=c^2+c-(u+1)(c+1)-(c+1)/c+u^2+u+2-2/c
=c^2+u^2-cu-3/c
=(u-c/2)^2+(3/4)c^2-3/c>=0,
记上式左边为f(u),当c/2>=2/√c，即c>=16^(1/3)时f(u)>=(3/4)c^2-3/c=3(c^3-4)/c>0,
当c/2<2/√c时f(u)>=f(2/√c)=c^2+4/c-2√c-3/c=c^2-2√c+1/c>=0.
∴原不等式成立。

Answer 2