已知a>0,B>0,C>0且a+b+c=1求证1/a+1/b+1/C>=9
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![已知a>0,B>0,C>0且a+b+c=1求证1/a+1/b+1/C>=9](/uploads/image/z/377750-38-0.jpg?t=%E5%B7%B2%E7%9F%A5a%3E0%2CB%3E0%2CC%3E0%E4%B8%94a%2Bb%2Bc%3D1%E6%B1%82%E8%AF%811%2Fa%2B1%2Fb%2B1%2FC%3E%3D9)
已知a>0,B>0,C>0且a+b+c=1求证1/a+1/b+1/C>=9
已知a>0,B>0,C>0且a+b+c=1求证1/a+1/b+1/C>=9
已知a>0,B>0,C>0且a+b+c=1求证1/a+1/b+1/C>=9
证明:∵a+b+c=1
∴1/a+1/b+1/c=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+b/a+c/a+a/b+1+c/b+a/c+b/c+1
=(b/a+a/b)+(c/a+a/c)+(b/c+c/b)+3
≥2+2+2+3=9
∴1/a+1/b+1/c≥9. 证毕!
1/a+1/b+1/
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=3+(b/a+a/b)+(c/a+a/c)+(b/c+c/b)
>=3+2+2+2=9
此时a=b=c=1/3
利用“1的替换”及均值不等式,解法如下:
1/a+1/b+1/c
=(a+b+c)/a+(a+b+c)/b+(a+b+c)/c
=1+b/a+c/a+a/b+1+c/b+a/c+b/c+1
=b/a+b/a+c/a+a/c+c/b+c/b+3
>=2+2+2+3=9
利用均值不等式,当且仅当1/a=1/b=1/c,即a=b=c=1/3时取等号;
证明:由题设及“柯西不等式”可得:1/a+1/b+1/c=(a+b+c)(1/a+1/b+1/c)≥(1+1+1)²=9.等号仅当a=b=c=1/3时取得。∴1/a+1/b+1/c≥9.