Question

一道不等式的高中数学竞赛题

Answer 1

第一问
首先，a b c为正数，则肯定1/（a+1）　1/（b+1）　1/（b+1）都小于1的．三个之和就小于
3,所以不等式前面一部分成立．
不等式后面一部分证法．
1/（a+1）+1/（b+1）+1/（b+1）≥3倍3次根号{1/[(a+1）(b+1)(c+1)]}
因为
(a+1）(b+1)(c+1)≤[(a+1)+(b+1)+(c+1)]^3/27=(a+b+c+3)^3/27≤(3+3)^3/27=8
则1/[(a+1）(b+1)(c+1)]≥1/8

所以
1/（a+1）+1/（b+1）+1/（b+1）≥3倍3次根号{1/[(a+1）(b+1)(c+1)]}≥3倍3次根号（1/8）＝3/2,当且仅当a=b=c=1时取等号

第二问
（a+1）/(a(a+2))=0.5[(2a+2)/(a(a+2)]=0.5[(a+(a+2))/(a(a+2)]=0.5[1/a+1/(a+2)]
所以原式＝0.5[（1/a+1/b+1/c)+1/(a+2)+(1/(a+2)+1/(b+2)+1/(c+2))]
利用第一问证明1/[(a+1）+1/(b+1)+1/(c+1)≥3/2的方法同样可以证得
1/a+1/b+1/c≥3
1/(a+2)+1/(b+2)+1/(c+2)≥1
所以原式＝0.5[（1/a+1/b+1/c)+1/(a+2)+(1/(a+2)+1/(b+2)+1/(c+2))]≥0.5（3+1）＝2
不等式成立

Answer 2

(1)根据柯西不等式得：
[(a+1)+(b+1)+(c+1)][1/(a+1)+1/(b+1)+1/(c+1)]>=(1+1+1)^2=9
即1/(a+1) + 1/(b+1) + 1/(c+1)>=9/(a+1+b+1+c+1)=9/(a+b+c+6)
又因为a+b+c<=3 故9/(a+b+c+6)>=9/3+3=3/2
所以1/(a+1) + 1/(b+1) + 1/(c+1)>=3/2
3-1/(a+1) + 1/(b+1) + 1/(c+1)=a/(a+1)+b/(b+1)+c/(c+1)
有因为a、b、c为正实数，所以a/(a+1)+b/(b+1)+c/(c+1)>0
故3-1/(a+1) + 1/(b+1) + 1/(c+1)>0<=>3>1/(a+1) + 1/(b+1) + 1/(c+1)
综合起来即为：3>1/(a+1) + 1/(b+1) + 1/(c+1)>=3/2
(2)因为a+1=1/2*(a+a+2)故原式可化为：
(a+1)/a(a+2) + (b+1)/b(b+2) + (c+1)/c(c+2)=1/2[1/a+1/(a+2) + 1/b+1/(b+2) + 1/c+1/(c+2)]
=1/2*[1/a +1/b+1/c+1/(a+2)+1/(b+2)+1/(c+2)]
同理根究柯西不等式可得：
1/a+1/b+1/c>=9/(a+b+c)>=9/3=3
1/(a+2)+1/(b+2)+1/(c+2)>=9/(a+2+b+2+c+2)=9/(a+b+c+6)>=9/(3+6)=1
故原式>=1/2*[3+1]=2 证毕！

Answer 3

柯西不等式一般形式 　　（∑（ai^2；））（∑（bi^2;)) ≥ （∑ai·bi)^2; 　　等号成立条件：a1:b1=a2:b2=…=an:bn，或ai、bi均为零。

根据柯西不等式得：
[(a+1)+(b+1)+(c+1)][1/(a+1)+1/(b+1)+1/(c+1)]>=(√[（a+1)*1/(a+1)]+√[（b+1)*1/(b+1)]+√[（c+1)*1/(c+1)])^2=(1+1+1)^2=9
即1/(a+1) + 1/(b+1) + 1/(c+1)>=9/(a+1+b+1+c+1)=9/(a+b+c+3)
又因为0＜a+b+c≤3 故3≤(a+b+c+3)≤6
3≥9/(a+b+c+3)0≥3/2
所以1/(a+1) + 1/(b+1) + 1/(c+1)>=3/2
(2)因为a+1=1/2*(a+a+2)故原式可化为：
(a+1)/a(a+2) + (b+1)/b(b+2) + (c+1)/c(c+2)=1/2[1/a+1/(a+2) + 1/b+1/(b+2) + 1/c+1/(c+2)]
=1/2*[1/a +1/b+1/c+1/(a+2)+1/(b+2)+1/(c+2)]
同理根究柯西不等式可得：
(a+b+c)(1/a+1/b+1/c)≥(1+1+1)^2=9
1/a+1/b+1/c=9/(a+b+c)
0＜a+b+c≤3
1/a+1/b+1/c=9/(a+b+c)≥3
同理
（a+2+b+2+c+2)[1/(a+2)+1/(b+2)+1/(c+2)]≥(1+1+1)^2=9

1/(a+2)+1/(b+2)+1/(c+2)>=9/(a+2+b+2+c+2)=9/(a+b+c+6)>=9/(3+6)=1
故原式>=1/2*[3+1]=2

Answer 4

晕