Question

急！·a,b∈R 求证(|a+b|)/(1+|a+b|)≤(|a|)/(1+|a|)+(|b|)/(1+|b|)

a,b∈R 求证(|a+b|)/(1+|a+b|)≤(|a|)/(1+|a|)+(|b|)/(1+|b|)
多谢！！！

Answer 1

千万别硬来。。。用倒数来证明就简单多了
先取倒数
1/[(|a|)/(1+|a|)+(|b|)/(1+|b|)]
=(1+|a|+|b|+|ab|)/(|a|+|b|+2|ab|)
=1+(1-|ab|)/(|a|+|b|+2|ab|)
而(1+|a+b|)/|a+b|=1+1/|a+b|

∵1-|ab|≤1
|a|+|b|+2|ab|≥|a+b|
∴(1-|ab|)/(|a|+|b|+2|ab|)≤1/|a+b|
1+(1-|ab|)/(|a|+|b|+2|ab|)≤1/|a+b|+1
即1/[(|a|)/(1+|a|)+(|b|)/(1+|b|)]≤(1+|a+b|)
∴(|a|)/(1+|a|)+(|b|)/(1+|b|)≥(|a+b|)/(1+|a+b|)

Answer 2

1/[(|a|)/(1+|a|)+(|b|)/(1+|b|)]
=(1+|a|+|b|+|ab|)/(|a|+|b|+2|ab|)
=1+(1-|ab|)/(|a|+|b|+2|ab|)
而(1+|a+b|)/|a+b|=1+1/|a+b|

∵1-|ab|≤1
|a|+|b|+2|ab|≥|a+b|
∴(1-|ab|)/(|a|+|b|+2|ab|)≤1/|a+b|
1+(1-|ab|)/(|a|+|b|+2|ab|)≤1/|a+b|+1
即1/[(|a|)/(1+|a|)+(|b|)/(1+|b|)]≤(1+|a+b|)
∴(|a|)/(1+|a|)+(|b|)/(1+|b|)≥(|a+b|)/(1+|a+b|)
!!!!!!!!!!!!!!

Answer 3

HARD

Answer 4

看不懂