For `a`, `b in R` prove that `(|a+b|)/(1+|a+b|) le (|a|)/(1+|a|)+(|b|)/(1+|b|)`

Explanation

If `a=0` or `b=0` or `a+b=0`, the inequality is obvious.
Let `ab(a+b) ne 0`
`(|a|)/(1+|a|)+(|b|)/(1+|b|) ge (|a|)/(1+|a|+|b|)+(|b|)/(|a|+1+|b|)=(|a|+|b|)/(1+|a|+|b|)=(1)/(1+(1)/(|a|+|b|))`………`(i)`
But `|a+b| le |a|+|b|`
`implies(1)/(|a+b|) ge (1)/(|a|+|b|)`
`implies1+(1)/(|a+b|) ge 1+(1)/(|a|+|b|)`
`implies (1)/(1+(1)/(|a+b|)) le (1)/(1+(1)/(|a|+|b|))`..........`(ii)`
From `(i)` and `(ii)` `implies (|a|)/(1+|a|)+(|b|)/(1+|b|) ge (1)/(1+(1)/(|a+b|))=(|a+b|)/(1+|a+b|)`