Question

已知a,b∈R a+b=1 求证ab+1/ab≧17/4

Answer 1

法一:
a,b∈R+,a+b=1≥2√ab→0≤ab≤1/4
设ab=x,则y=ab+1/ab=x+1/x.
对于函数f(x)=y=x+1/x,f(x)在(-∞,-1)或(1,+∞)上单调递增,f(x)在(-1,0)或(0,1)上单调递减(可用函数单调定义证,也可求导证)
∴ab=1/4时y=ab+1/ab取最小值,y最小=4+(1/4)=17/4

法二:
a,b∈R+,a+b=1≥2√ab→ab≤1/4→1/ab≥4
y=ab+1/ab=ab+1/16ab+15/16ab≥2√(ab*1/16ab)+15/16ab≥1/2+(15/16)*4=17/4.当且仅当a=b=1/2时,y取最小值17/4

跪求采纳，日子不好过啊