(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]=(x^2+y^2+x^2*y^2+1)/xy=x/y+y/x+xy+1/xy (xy+1/xy不能用均值定理）=x/y+y/x+xy+（x+y)^2/xy=2(x/y+y/x)+xy+2 (1=x+y≥2√xy),xy≤1/4，)≥6+xy=6.25此時x=y=1/2方法2(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]=(x^2+y^2+x^2*y^2+1)/xy=[(x+y)^2-2xy+(xy)^2+1]/xy=[2-2xy+(xy)^2]/xy=2/xy+xy-2.設t=xy≤[(x+y)/2]^2=1/4.f(t)=2/t+t在（0，√2）單減，在（√2，+∞）單增。f(t)=2/t+t在t=1/4時取得最小值。代入得最小為25/42）解：因a>b>0.故a²＞ab＞0.===>a²-ab＞0,且ab＞0.由基本不等式可知；a²+(1/ab)+[1/(a²-ab)]={(a²-ab)+[1/(a²-ab)]}+[(ab)+1/(ab)]≥2+2=4。等號僅當a²-ab=1,ab=1時取得；即當a=√2,b=1/√2時取得。故原式min=4.
(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]=(x^2+y^2+x^2*y^2+1)/xy=x/y+y/x+xy+1/xy (xy+1/xy不能用均值定理）=x/y+y/x+xy+（x+y)^2/xy=2(x/y+y/x)+xy+2 (1=x+y≥2√xy),xy≤1/4，)≥6+xy=6.25此時x=y=1/2方法2(x+1/x)(y+1/y)=[(x^2+1)/x][(y^2+1)/y]=(x^2+y^2+x^2*y^2+1)/xy=[(x+y)^2-2xy+(xy)^2+1]/xy=[2-2xy+(xy)^2]/xy=2/xy+xy-2.設t=xy≤[(x+y)/2]^2=1/4.f(t)=2/t+t在（0，√2）單減，在（√2，+∞）單增。f(t)=2/t+t在t=1/4時取得最小值。代入得最小為25/42）解：因a>b>0.故a²＞ab＞0.===>a²-ab＞0,且ab＞0.由基本不等式可知；a²+(1/ab)+[1/(a²-ab)]={(a²-ab)+[1/(a²-ab)]}+[(ab)+1/(ab)]≥2+2=4。等號僅當a²-ab=1,ab=1時取得；即當a=√2,b=1/√2時取得。故原式min=4.