若正数a,b满足(1/a)+(1/b)=1,则[4/(a-1)]+[16/(b-1)]的最小值为( )求详解,

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若正数a,b满足(1/a)+(1/b)=1,则[4/(a-1)]+[16/(b-1)]的最小值为( )求详解,
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若正数a,b满足(1/a)+(1/b)=1,则[4/(a-1)]+[16/(b-1)]的最小值为( )求详解,
若正数a,b满足(1/a)+(1/b)=1,则[4/(a-1)]+[16/(b-1)]的最小值为( )求详解,

若正数a,b满足(1/a)+(1/b)=1,则[4/(a-1)]+[16/(b-1)]的最小值为( )求详解,
1/a+1/b=1,
∴ab-a-b=0且b=a/(a-1).
∴设4/(a-1)+16/(b-1)=t
→16a+4b-20=t(ab-a-b+1)
→t=16a+4b-20
→t=16a+[4a/(a-1)]-20
→t=16(a-1)+[4/(a-1)].
故依基本不等式得
t≥2√[16(a-1)·4/(a-1)]=16.
取等时,16(a-1)=4/(a-1),
即a=3/2,b=3.
故a=3/2,b=3时,
所求最小值为:16.