# What is maxim of summatio? maxim(1-a,b+c)+maxim(1-c,a+b)+maxim(1-b,a+c)?a ,b,c real numbers

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You need to use the definition of maximum function such that:

`max(a,b) = {(a, if a>b),(b, if a < b):}`

Reasoning by analogy yields:

`max(1-a,b+c) = {(1 - a, if 1 - a > b + c), (b + c, if 1 - a < b + c):}`

`max(1-c,a+b) = {(1 - c, if1 - c > a + b), (a + b, if 1 - c< a + b):}`

`max(1-b,a+c) = {(1 - b, if 1 - b > a + c), (a + c, if 1 -b < a + c):}`

You need to consider `max(1-a,b+c) = 1 - a, max(1-c,a+b) = 1 - c` and `max(1-b,a+c) = 1 - b` such that:

`1 - a + 1 - b + 1 - c = 3 - (a + b + c)`

You need to consider `max(1-a,b+c) = 1 - a, max(1-c,a+b) = a + b` and `max(1-b,a+c) = 1 - b` such that:

`1 - a + a + b + 1 - b = 2`

You need to consider `max(1-a,b+c) = 1 - a, max(1-c,a+b) = 1 - c` and `max(1-b,a+c) =a + c` such that:

`1 - a + 1 - c + a + c = 2`

You need to consider `max(1-a,b+c) = b + c, max(1-c,a+b) = 1 - c` and `max(1-b,a+c) =1 - b` such that:

`b + c + 1 - c + 1 - b = 2`

You need to consider `max(1-a,b+c) = b + c, max(1-c,a+b) =a + b` and `max(1-b,a+c) = 1 - b ` such that:

`b + c + a + b + 1 - b = 1 + a + b + c`

You need to consider `max(1-a,b+c) = b + c, max(1-c,a+b) =1 - c` and `max(1-b,a+c) =a + c` such that:

`b + c + 1 - c + a + c = 1 + a + b + c`

You need to consider `max(1-a,b+c) = b + c, max(1-c,a+b) =a + b` and `max(1-b,a+c) = a + c ` such that:

`b + c + a + b + a + c = 2(a + b + c)`

**Hence, evaluating the maximum sums, under different conditions yields the following values such that: `3 - (a + b + c), 2, 2(a + b + c) and 1 + a + b + c.` **