已知a>0,b>0,c>0求证：（a+b+c）[(1/a)+(1/b)+(1/c)≥9
热心网友
原始=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c=1+b/a+c/a+1+a/b+c/b+1+b/c+a/c=3+(a/b+b/a)+(b/c+c/b)+(a/c+c/a)因为a0,b0,c0所以a/b+b/a≥2√(a/b)(b/a)=2b/c+c/b≥2√(b/c)(c/b)=2,a/c+c/a≥2√(a/c)(c/a)=2所以原式≥3+2+2+3=9,得证
热心网友
由柯西不等式得（a+b+c）[(1/a)+(1/b)+(1/c)]=(a/a+b/b+c/c)^2=9