Question

已知f(x)=m-(x-2),m属于R,f(x+2)≥0的解集为[-1,1],求证若a,b,c属于R,且1/a+1/2b+1/3c=m

求证a+2b+3c≥9

Answer 1

f(x+2)=m-|x| f（x+2）≥0的解集为【-1，1】.
即 |x|-m<=0 |x|<=m 解集为【-1，1】. 所以m=1
1/a+1/2b+1/3c=1
a+2b+3c
=(a+2b+3c)(1/a+1/2b+1/3c)
=1+1+1+a/2b+2b/a+a/3c+3c/a+2b/3c+3c/2b
=3+(a/2b+2b/a)+(a/3c+3c/a)+(2b/3c+3c/2b)
>=3+2+2+2
=9