已知a>0,b>0,且a+b=1.求证

左式=ab+a/b+1/ab+b/a
=(a2b2+a2+1+b2)/ab
=[a2b2+(1-2ab)+1]/ab
=[(ab-1)2+1]/ab

a+b=1
ab<=[(a+b)/2]²=1/4

所以(ab-1)^2+1≥25/16,0<ab≤1/4,1/ab≥4
相乘得到，左式≥25/4

========================================================
公理：（a+b）^2/4 <= a^2+b^2

根号下a+1/2 +根号下b+1/2
≤根号(4*(a+b+1))
<根号(4*(1+1))
<根号8
≤ 2

----------------------------------------------------------
a+b>=2根号下ab
ab<=1/4
1/2(a+b)+1/4+ab<=1
(a+1/2)(b+1/2)<=1
2*根号下【(a+1/2)(b+1/2)】<=2
a+b+1+2*根号下【(a+1/2)(b+1/2)】<=4
[根号下(a+1/2)+根号下（b+1/2)]^2<=4
根号下(a+1/2)+根号下（b+1/2)<=2