1、课时规范练59不等式的证明基础巩固组1.(2020全国,理23)设a,b,cR,a+b+c=0,abc=1.(1)证明:ab+bc+ca0;(2)用maxa,b,c表示a,b,c的最大值,证明:maxa,b,c34.证明(1)由题设可知,a,b,c均不为零,所以ab+bc+ca=12(a+b+c)2-(a2+b2+c2)=-12(a2+b2+c2)0,b0,c0,b0且a2+b2=2.(1)若使1a2+4b2|2x-1|-|x-1|恒成立,求x的取值范围;(2)证明:1a+1b(a5+b5)4.(1)解a,b(0,+),且a2+b2=2,1a2+4b2=12(a2+b2)1a2+4b2=121
2、+4+b2a2+4a2b2125+2b2a24a2b2=92,当且仅当b2=2a2时,等号成立,则|2x-1|-|x-1|92,当x12时,不等式化为1-2x+x-192,解得-92x12;当12x1时,不等式化为2x-1+x-192,解得12x1;当x1时,不等式化为2x-1-x+192,解得1x92.综上,x的取值范围为-92,92.(2)证明(方法1)1a+1b(a5+b5)=a4+b4+a5b+b5a=(a2+b2)2-2a2b2+a5b+b5a4-2a2b2+2a5bb5a=4-2a2b2+2a2b2=4,当且仅当a=b=1时,等号成立.(方法2)由柯西不等式可得1a+1b(a5+b
3、5)=1a2+1b2(a52)2+(b52)2a52a+b52b2=(a2+b2)2=4,当且仅当a=b=1时,等号成立.综合提升组4.(2021江西赣县模拟)已知f(x)=|x+1|+|x-3|.(1)求不等式f(x)x+3的解集;(2)若f(x)的最小值为m,正实数a,b,c满足a+b+c=m,求证:1a+b+1b+c+1a+c92m.(1)解当x-1时,2-2xx+3,解得x-13,则不等式的解集为空集;当-13时,2x-2x+3,解得x5,则31时,得2x-14,解得x52;当0x1时,得14,此时不等式无解;当x0).(1)当m=1时,求不等式f(x)6的解集;(2)若f(x)的最小
4、值为32,且a+b=m(a0,b0),求证:a+2b5.(1)解当m=1时,原不等式为|x+1|+|2x-1|6,则x12,x+1+2x-16,解得-2x-1或-1x12或12x2,原不等式f(x)6的解集为x|-2x2.(2)证明由题意得f(x)=-3x-m2+m,xm2,f(x)min=fm2=m2+12m=32,m=1或m=-32(舍去),a+b=1,令a=cos2,b=sin202,则a+2b=cos+2sin=5sin(+)5,当=2-0bc,a2+b2+c2=1,求证:a+b1.证明(1)c=0时,a+b=1,a+1a2+b+1b2(a+1a)2+(b+1b)2+2(a+1a)(b
5、+1b)2=(a+1a)+(b+1b)22=(1+1a+1b)22,a,bR+,a+b=1,1a+1b=1a+1b(a+b)=2+ba+ab2+2baab=4,从而a+1a2+b+1b2(1+4)22=252,当且仅当a+b=1,a+1a=b+1b,ba=ab,即a=b=12时,等号成立;(2)假设a+b1,则由a+b+c=1,知c0,故abc0,又由(a+b+c)2=a2+b2+c2+2ab+2bc+2ac=1,得ab+bc+ac=0,但由abc0,知ab+bc+ac0,矛盾,故假设a+b1不成立,则a+b1.8.(2021陕西宝鸡二模)已知函数f(x)=|2x-4|+|x+1|.(1)求不
6、等式f(x)8的解集;(2)设a,b,cR,且a+b+c=1,证明:a3bc+b3ac+c3ab1.(1)解由题意得,f(x)=|2x-4|+|x+1|=3-3x,x2,不等式f(x)8,可转化为3-3x8,x2,解得x-53或x113,故不等式的解集为xx-53或x113.(2)证明a4+b42a2b2,b4+c42b2c2,c4+a42c2a2,三式相加得a4+b4+c4a2b2+b2c2+c2a2,又因为a2b2+b2c22ab2c,a2b2+c2a22a2bc,b2c2+c2a22abc2,三式相加得a2b2+b2c2+c2a2abc(a+b+c),又因为a+b+c=1,所以a2b2+b2c2+c2a2abc,即a4+b4+c4abc,又因为abc0,所以a4+b4+c4abc1,即a3bc+b3ac+c3ab1,当且仅当a=b=c=13时,等号成立.
