高考第一輪文科數(shù)學(xué)（人教A版）課時(shí)規(guī)范練59　不等式的證明

課時(shí)規(guī)范練59不等式的證明基礎(chǔ)鞏固組1.(2020全國(guó)Ⅲ,文23)設(shè)a,b,c∈R,a+b+c=0,abc=1.(1)證明:ab+bc+ca<0;(2)用max{a,b,c}表示a,b,c的最大值,證明:max{a,b,c}≥342.(2022陜西西安二模)已知函數(shù)f(x)=|x-4|+|x-2|.(1)求不等式f(x)≤4的解集;(2)若函數(shù)f(x)的最小值為m,正實(shí)數(shù)a,b滿足a+b=m,求證:1a+2+3.(2022安徽安慶二模)已知函數(shù)f(x)=|2x+4|+|x-1|.(1)求不等式f(x)>6的解集;(2)設(shè)函數(shù)f(x)的最小值為m,正實(shí)數(shù)a,b滿足a2+9b2=m,求證:a+3b≥26ab.綜合提升組4.已知f(x)=|x+1|+|x-3|.(1)求不等式f(x)≤x+3的解集;(2)若f(x)的最小值為m,正實(shí)數(shù)a,b,c滿足a+b+c=m,求證:1a5.已知函數(shù)f(x)=|x-1|.(1)解不等式f(x)+f(x+1)≥4;(2)當(dāng)x≠0,x∈R時(shí),證明:f(-x)+f1x≥2.6.已知函數(shù)f(x)=|x+m2|+|2x-m|(m>0).(1)當(dāng)m=1時(shí),求不等式f(x)≤6的解集;(2)若f(x)的最小值為32,且a+b=m(a>0,b>0),求證:a+2b創(chuàng)新應(yīng)用組7.(2022江西南昌三模)已知函數(shù)f(x)=|x-2|+|x-4|,不等式f(x)≥kx(k>0)恒成立.(1)求k的最大值k0;(2)設(shè)a>0,b>0,求證:aa8.已知函數(shù)f(x)=|2x-4|+|x+1|.(1)求不等式f(x)≥8的解集;(2)設(shè)a,b,c∈R,且a+b+c=1,證明:a3bc

參考答案課時(shí)規(guī)范練59不等式的證明1.證明(1)由題設(shè)可知,a,b,c均不為零,所以ab+bc+ca=12[(a+b+c)2-(a2+b2+c2)]=-12(a2+b2+c2)<(2)不妨設(shè)max{a,b,c}=a,因?yàn)閍bc=1,a=-(b+c),所以a>0,b<0,c<0.由bc≤(b+c)24,可得abc≤a34,故a≥34,所以2.(1)解當(dāng)x<2時(shí),f(x)=4-x+2-x=6-2x≤4,解得1≤x<2;當(dāng)2≤x≤4時(shí),f(x)=4-x+x-2=2≤4,解得2≤x≤4;當(dāng)x>4時(shí),f(x)=x-4+x-2=2x-6≤4,解得4<x≤5.綜上,不等式f(x)≤4的解集為[1,5].(2)證明由(1)得,當(dāng)x<2時(shí),f(x)=6-2x>2;當(dāng)2≤x≤4時(shí),f(x)=2;當(dāng)x>4時(shí),f(x)=2x-6>2,所以a+b=m=2.又a,b為正實(shí)數(shù),所以0<b<2,故1a+2+13.解(1)由條件可知原不等式可化為①x②-③x解①得x>1;解②得x∈?;解③得x<-3,所以原不等式的解集為(-∞,-3)∪(1,+∞).(2)因?yàn)閒(x)=|2x+4|+|x-1|=3所以當(dāng)x=-2時(shí),函數(shù)f(x)的最小值為m=3,于是a2+9b2=3,∵a>0,b>0,而3=a2+9b2≥2a·3b=6ab,于是0<ab≤12∵a+3bab=1∴a+3b≥26ab,原不等式得證.4.(1)解①當(dāng)x≤-1時(shí),2-2x≤x+3,解得x≥-13,則不等式的解集為空集②當(dāng)-1<x≤3時(shí),4≤x+3,解得1≤x≤3;③當(dāng)x>3時(shí),2x-2≤x+3,解得x≤5,則3<x≤5.綜上,不等式的解集為{x|1≤x≤5}.(2)證明因?yàn)閒(x)=|x+1|+|x-3|≥|x+1-x+3|=4,當(dāng)且僅當(dāng)(x+1)·(x-3)≤0時(shí),等號(hào)成立.所以m=4,所以a+b+c=m=4,1a+b+1b+c+1c+a=18[(a+b)+(b+c)+(c+a)]·1a+b+1b+c+1c+a=5.(1)解由f(x)+f(x+1)≥4得|x-1|+|x|≥4,當(dāng)x>1時(shí),得2x-1≥4,解得x≥52當(dāng)0≤x≤1時(shí),得1≥4,此時(shí)不等式無解;當(dāng)x<0時(shí),得-2x+1≥4,此時(shí)x≤-32所以不等式的解集為xx≥52或x≤-32.(2)證明f(-x)+f1x=|x+1|+1x-1,由絕對(duì)值三角不等式,得|x+1|+1x-1≥x+1x,又因?yàn)閤,1x同號(hào),所以x+1x=|x|+1x,由基本不等式得|x|+1x≥2,當(dāng)且僅當(dāng)|x|=1時(shí),等號(hào)成立,所以f(-x)+f1x≥2.6.(1)解當(dāng)m=1時(shí),原不等式為|x+1|+|2x-1|≤6,則x或-或x解得-2≤x<-1或-1≤x≤12或12<x≤2,∴原不等式f(x)≤6的解集為{x|-(2)證明由題意得f(x)=-∴f(x)min=fm2=m2+12m=32,∴m=1或m=-32(∴a+b=1,令a=cos2θ,b=sin2θ0<θ<π2,則a+2b=cosθ+2sinθ當(dāng)θ=π2-φ0<φ<π2,且tanφ=12時(shí),上述不等式等號(hào)成立7.(1)解當(dāng)x≤2時(shí),f(x)=2-x+4-x=6-2x;當(dāng)2<x<4時(shí),f(x)=x-2+4-x=2;當(dāng)x≥4時(shí),f(x)=x-2+x-4=2x-6.由此可得f(x)的圖象如下圖所示,∵f(x)≥kx(k>0)恒成立,則由圖象可知,當(dāng)y=kx過點(diǎn)(4,2)時(shí),k取得最大值k0,∴k0=12(2)證明由(1)知,只需證明aa令m解得a=2n-m3,b=2m-n3當(dāng)且僅當(dāng)2nm=2mn,即m=n時(shí),等號(hào)成立,8.(1)解由題意得,f(x)=|2x-4|+|x+1|=3不等式f(x)≥8,可轉(zhuǎn)化為3解得x≤-53或x≥11故不等式的解集為xx≤-53或x≥113.(2)證明a4+b4≥2a2b2,b4+c4≥2b2c2,c4+a4≥2c2a2,三式相加得a4+b4+c4≥a2b2+b2c2+c2a2,又因?yàn)閍2b2+b2c2≥2ab2c,a2b2+c2a2