In Figure, prove that `C D+D A+A B+B C >2A C` `C D+D A+A B+B C >2A C`

Explanation

Solution
'(i)' Consider'/_\CDA'
We know that 'CD+DA>AC..(1)'
Consider '/_\ABC'
We know that 'AC+AB>BC..(2)'
By adding both the equations, we get
'CD+DA+AC+AB>AC+BC'
By subtracting 'AC' on both the sides
'CD+DA+AC+ABAC>AC+BCAC'
So we get
'CD+DA+AB>BC
Therefore, it is proved that 'CD+DA+AB>BC'
'(ii)' Consider '/_\CDA'
We know that 'CD+DA>AC..(1) '
Consider '/_\ABC'
We know that 'AB+BC>AC..(2)'
By adding both the equations we get
'CD+DA+AB+BC>AC+AC'
So we get 'CD+DA+AB+BC>2AC'
Therefore, it is proved that 'CD+DA+AB+BC>2AC'.