不等式填空题:
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不等式填空题:
不等式填空题:
不等式填空题:
请参考下面的答案:
1/(3a+2) +1/(3b+2) +1/(3c+2)
=[(3a+2)/9+1/(3a+2)] +[(3b+2)/9+1/(3b+2)] +[(3c+2)/9+1/(3c+2)] - 1
≥2√[(3a+2)/9·1/(3a+2)] +2√[(3b+2)/9·1/(3b+2)] +2√[(3c+2)/9·1/(3c+2)] -1
=2/3 +2/3 +2/3 -1
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1/(3a+2) +1/(3b+2) +1/(3c+2)
=[(3a+2)/9+1/(3a+2)] +[(3b+2)/9+1/(3b+2)] +[(3c+2)/9+1/(3c+2)] - 1
≥2√[(3a+2)/9·1/(3a+2)] +2√[(3b+2)/9·1/(3b+2)] +2√[(3c+2)/9·1/(3c+2)] -1
=2/3 +2/3 +2/3 -1
=1
从而当且仅当(3a+2)/9=1/(3a+2),(3b+2)/9=1/(3b+2), (3c+2)/9=1/(3c+2)
即 a=b=c=1/3时,原式有最小值为1
收起
这个根据不等式链的变式啊,3/(1/a+1/b+1/c)<=(abc)^1/3<=(a+b+c)/3
这里只要稍微变换一下形式,令 A=3a+2;B=3b+2;C=3c+2
则a+b+c=(A+B+C)/3-2=1; (A+B+C)/3=3
原题变形为已知 (A+B+C)/3=3 求1/A+1/b+1/C的最小值
由不等式链可得3/(1/A+1/B+1/C)<=(A+B+C)/3=3 所以(1/A+1/B+1/C)>=1