Question

已知a,b,c为 实数 ,且a+b+c-2-2m=0,a^2+1/4b^2+1/9c^2+m-1=0

已知a,b,c为 实数 ,且a+b+c-2-2m=0,a^2+1/4b^2+1/9c^2+m-1=0.
1.求证a^2+1/4b^2+1/9c^2>=（a+b+c）^2/14
2.求实数m的取值范围。
是a+b+c+2-2m=0,不是a+b+c-2-2m=0

Answer 1

a^2+1/4b^2+1/9c^2=1-m
a+b+c=2m-2 （a+b+c）^2/14=4（m+1）^2/14

a^2+1/4b^2+1/9c^2-（a+b+c）^2/14
=1-m -（m-1））^2 / 7
=-2m^2-3m+5≥0 则求出 -5/2≤m≤1

Answer 2

a+b+c+2-2m=a^2+1/4b^2+1/9c^2+m-1
a+b+c=a^2+1/4b^2+1/9c^2+3m-3
(a+b+c）^2/14=(a^2+1/4b^2+1/9c^2+3m-3)^2/14

若m=0，a+b+c=-2，a^2+1/4b^2+1/9c^2=1
a^2+1/4b^2+1/9c^2>（a+b+c）^2/14
若m=1,a^2+1/4b^2+1/9c^2=（a+b+c）^2/14
若m<0，a^2+1/4b^2+1/9c^2>(a+b+c）^2/14
若m>1，a^2+1/4b^2+1/9c^2>(a+b+c）^2/14
所以a^2+1/4b^2+1/9c^2>=（a+b+c）^2/14