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已知a,b是正数,且满足2<a+2b<4,那么b+1a+1的...

如图所示,画出可行域,2<a+2b<4a>0b>0,b+1a+1表示可行域内的点Q(a,b)与P(-1,-1)所在直线的斜率.A(4,0),B(0,2)而kPA=?1?0?1?4=15,kPB=?1?2?1?0=3.∴15<b+1a+1<3.故答案为:(15,3).

∵正实数a,b满足1a+2b=3,∴3≥21a?2b,化为ab≥89,当且仅当b=2a=43时取等号.b+2a=3ab.∴(a+1)(b+2)=ab+b+2a+2=4ab+2≥329+2=509.故答案为:509.

∵正实数a,b,c满足a+2b+c=1,令a+b=x>0,b+c=y>0,且x+y=1.∴1a+b+9(a+b)b+c=1x+9xy,由x+y=1可得y=1-x.∴1x+9xy=1x+9x1?x=f(x).(0<x<1)∴f′(x)=-1x2+9(1?x)?9x?(?1)(1?x)2=8x2+2x?1x2(1?x)2=(2x+1)(4x?1)x2(1?x)2,令f′(x)=0,解...

1a+2b=(1a+2b)(a+2b)=1+2ba+2ab+4≥5+22ba?2ab=9.当且仅当2ba=2ab,即a=b=13时,取最小值9.

根据平均值不等式1a+1b≥2ab,∴1a+1b+2ab≥2ab+2ab≥22ab×2ab=4.故答案为:4.

a>0,b>0,且a+b=1,令ab=t,则 由 1=(a+b)2=a2+b2+2ab≥4ab,可得 0<ab≤14,则 1a+1b+2ab=1t2+2t,t∈(0,12],而函数y=1t2+2t,则y′=2?2t3<0,则当t=12时,1a+1b+2ab取最小值5.故选D.

解:因为已知a+b=1,a>0,b>0, ∴根据基本不等式a+b≥2 √ab , ∴0<ab≤ 14 , 又(a+ 1a )(b+ 1b )= a2+1a ⋅ b2+1b = a2b2-2ab+2ab = (1-ab)2+1ab ≥ 254 (取等号时a=b= 12 ) ∴(a+ 1a )(b+ 1b )≥ 254 即得(a+ 1a )(b+ 1b )≥ 254 .

证明:∵a+b+c=1,∴(1a?1)(1b?1)(1c?1)=b+ca×a+cb×a+bc∵a,b,c∈R+,∴b+ca×a+cb×a+bc≥2bca×2acb×2abc=8∵1π∫4?416?x2dx=1π×12π×42=8∴(1a?1)(1b?1)(1c?1)≥1π∫4?4

(1)∵2ab=a2+b2≥2ab,即ab≥ab,∴ab≤1.又∴1a+1b≥2ab≥2,当且仅当a=b时取等号.∴m=2.(2)函数f(x)=|x-t|+|x+1t|≥|t+1t|≥2>22=1,∴满足条件的实数x不存在.

解答:证明:(1)∵a+b+c=1,∴1a+1b+1c=(a+b+c)(1a+1b+1c)=3+(ba+ab)+(ca+ac)+(cb+bc),∵a、b、c均为正数,∴ba+ab≥2,ca+ac≥2,cb+bc≥2,代入上式,得1a+1b+1c≥9 (2)∵a,b,c均为正数,∴a2+b2≥2ab,a2+c2≥2ac,b2+c2≥2bc,以上三式...

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