Question

13.已知a>b>c >0,且2a^2+1/ab+1/a(a-b)-4ac+4c^2=4 ,则a+b+c= ▲ 注；b大

Answer 1

2a^2+1/ab+1/a(a-b)-4ac+4c^2=a^2+1/ab+1/a(a-b)+(a-2c）^2

》a^2+1/ab+1/a(a-b)=a^2+1/b(a-b)=[(a-b)-b]^2+1/b(a-b)=(a-b)^2+b^2-2(a-b)b+1/b(a-b)

》2(a-b)b+2(a-b)b+1/b(a-b)=4(a-b)b+1/b(a-b)》4 ,所以最小值是4

当且仅当a-b=b且a=2c时,4(a-b)b=1/b(a-b时取等号

此时a=根号2 b=c=根号2/2

a+b+c=2*根下2