Consider the two expressions, (a + 1)² and (a + 1)³.
No matter what the number is, a² ≥ 0. Even if you square a negative number, you still get a positive number. We include the "or equal to" since 0² = 0. Since a² ≥ 0 is always true, then we can manipulate this inequality.
a² ≥ 0
a² + 2a + 1 ≥ 0 + 2a +1 by adding 2a +1 to both sides
(a + 1)² ≥ 2a + 1 as a² + 2a + 1 factors into (a +1)²
(a + 1)² ≥ 1 If it's more than 2a +1, it has to be more than 1
(a + 1)² ≥ 0 If it's more than one, it's more than zero since 1 > 0.
So we can conclude that (a + 1) ≥ 0 is always true.
Now let's look at (a + 1)³. We know that from before (a + 1)² ≥ 0. A tempting thing to say is that if you multiply by both sides by (a + 1) then both will be more than zero. Doing so isn't right.
What we instead must do is use cases. Either a is positive, negative, or zero.
Case 1: a = 0
When a = 0, (a + 1)³ = 1³ = 1 and 1 > 0. Thus (a + 1)³ ≥ 0,
Case 2: a > 0
When a > 0, a² > 0 from multiplying both sides by a. Do it a second time and a³ > 0. Then, if we add terms to both sides like in the a² example, we have this:
a³ > 0
a³ + 3a² > 3a² by adding 3a² to both sides
a³ + 3a² + 3a > 3a² + 3a by adding 3a to both sides
a³ + 3a² + 3a + 1 > 3a² + 3a + 1 by adding 1 to both sides
(a + 1)³ > 3a² + 3a + 1 by factoring the left side
(a + 1)³ > 1 Since a > 0 by assumption then 3a > 0 and 3a² > 0 and their sum is more than zero too
(a + 1)³ > 0 Since 1 > 0
Case 3: a < 0
Since a < 0 then a² > 0 (minus times minus is plus), but a³ < 0 through a similar multiplication.
Let b = a + 1. If a < 0 then a + 1 < 1 by adding 1 to both sides and b < 1 by back substitution. So b³ < 1³ by cubing both sides and b³ < 1 since 1³ =1.
b³ < 1
(a+1)³ < 1 We chose a + 1 = b.
When a < 0, we can conclude that (a + 1)³ < 1. When a ≥ 0, then (a +1)³ ≥ 0.
However for all a, (a +1)² ≥ 0. Thus, we have our sometimes truth. That when we choose a to be negative we have that the (a + 1)² and (a + 1)³ are of opposite signs.
