Proof by working backwards(3)

The expression (x^4)/4 + (x + 1)^3 > ((x +1)^4)/4 for all real numbers x >= -1.

(Scratch work)

x^4 + 4(x^2 +2x +1)(x +1) > (x +1)^2(x^2 +2x +1)

x^4 + 4(x^3 +2x^2 +x +x^2 +2x +1) > (x+1)(x^3 +3x^2 +3x +1)

x^4 + 4(x^3 +3x^2 +3x +1) > x^4 +4x^3 +6x^2 +4x +1

x^4 + 4x^3 +12x^2 + 12x + 4 > x^4 + 4x^3 + 6x^2 + 4x + 1

6x^2 + 6x + 3 > 0

3x^2 +3x +1 > 0

Proof. Let f(x) = 3x^2 + 3x + 1, then f '(x) = 6x + 3 which implies that f attains its minimum value at x = -1/2. So for all x > -1,

f(x) >= 3/4 - 3/2 +1 = 1/4 > 0

Therefore we have that

3x^2 +3x +1 > 0 =>

6x^2 + 6x +3 > 0 =>

x^4 +  4x^3 +12x^2 + 12x +4 > x^4 +4x^3 +6x^2 +4x +1 =>

x^4 + 4(x^3 +3x^2 +3x +1) > x^4 +4x^3 +6x^2 +4x +1 =>

x^4 +4(x^3 +2x^2 + x + x^2 + 2x +1) > (x + 1)(x^3 + 3x^2 + 3x +1) =>

x^4 + 4(x^2 +2x + 1)(x + 1) > (x + 1)^2 (x^2 +2x + 1) =>

(x^4)/4 + (x + 1)^3 > ((x + 1)^4)/4

QED

Proof by working backwards(2)

(2) If n^3 + 5n + 6 is divisible by 3 for some integer n, then (n + 1)^3 + 5(n + 1) + 6 is divisible by 3.

(Scratch work)

(n + 1)^3 + 5(n + 1) + 6 = (n + 1)(n^2 + 2n +1) +5n + 5 + 6
                                     
                                       = n^3 + 2n^2 + n + n^2 + 2n + 1 + 5n + 11

                                       = n^3 + 5n + 6 + 3n^2 + 3n + 6

                                       = n^3 + 5n + 6 + 3(n^2 + n + 2)


Proof. Assume that n^3 + 5n + 6 is divisible by 3 for some integer n. Then

3 | (n^3 + 5n + 6) =>

3 | [(n^3 + 5n + 6) + 3(n^2 +n + 2)] =>

3 | [n^3 + 5n + 6 + 3n^2 + 3n + 6] =>

3| [n^3 + 2n^2 +n + n^2 + 2n + 1 + 5n + 11] =>

3| [(n + 1)(n^2 + 2n + 1) + 5n +5 +6] =>

3| [(n + 1)^3 + 5(n + 1) + 6]

QED

Proof by working backwards(1)

Write clear and complete proofs by working backwards for the following mathematical statement.
(1) The expression x + 9/x >= for all real numbers x > 0.
(Scratch work)

x + 9/x >= 6

x^2 + 9 >= 6x

x^2 - 6x + 9 >= 0

(x - 3)^2 >= 0

Proof. Assume that x is a real number with x > 0. Then notice that

(x -3)^2 >= 0

x^2 - 6x +9 >= 0

x^2 + 9 >= 6x

x + 9/x >= 6x.

QED

Proof by cases(4)

If a and b  are real numbers.

Case 1: Assume that a >= 0, b >= 0. Then |a| = a, |b| = b, and |a - b| = a - b. So in this case
||a| - |b|| = |a - b| =< |a - b|.

Case 2: Assume that a >= 0, and b < 0. Then |a| = a and |b| = -b.

-Sub-case 1: Assume that |a| >= |b|, then 0=< a + b =< a =< a - b.
So |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|.

-Sub-case 2: Assume that |b| > |a|, then we have both 0 < -b =< a - b, and b < a + b < 0. So it follows that |a + b| < |a - b| giving us

||a| - |b|| = |a + b| < |a - b|

Case 3: Assume that a < 0, and b >= 0. Then |a| = - a and |b| = b.
-Sub-case 1: Assume that |a| >= |b|, then 0 < - a =< b - a = |a - b|, and a=< a + b =< 0. So that |a + b| =< |a - b|. It follows that

||a| - |b|| = |- a - b| = |a + b| =< |a - b|.

-Sub-case 2: Assume that |b| > |a|, then a - b =< - a - b< 0. This gives us |- a - b| =< |a - b|. So again
||a| - |b|| = |- a - b| =< |a - b|.

Case 4: Assume that a < 0, and b < 0. Then |a| = - a and |b| = - b.
Then in this case
||a| - |b|| = |- a + b| = |a - b| =< |a - b|.

So in all cases ||a| - |b|| =< |a - b|

QED

Proofs by Case (3)

(3) If n is an even integer, then n = 4j or n = 4j - 2 for some integer j.

Proof. Assume that n is an even integer. Then n = 2k for some integer k.
Case 1: Assume that k is even. Then k = 2p for some integer p. So in this case

n = 2k = 2(2p) = 4p

and the results holds.

Case 2: Assume that k is odd. Then k = 2p + 1 for some integer p. So in this case

n = 2k = 2(2p + 1) = 4p + 2 + 4 - 4 = 4(p - 1) - 2 = 4j - 2

where j = p - 1 is an integer, and therefore the result holds.

So in all cases n = 4j or n = 4j - 2 for some integer j.

QED

Proofs by Case (2)

Write clear and complete proofs by cases for the mathematical statement.

(2) If x is a real number, then |x - 1| + |x + 5| >= 6.

Proof. Assume that x is a real number. Consider the following cases:

Case 1: Assume that x  < -5. In this case x - 1 < 0, and x + 5 < 0. So |x -1| = -(x - 1) = 1 - x, and |x + 5| = -(x + 5) = -5 - x. Then

|x - 1| + |x + 5| = 1 - x - 5 - x = -4 - 2x > -4 -2(-5) = -4 + 10 = 6.

So in this case |x - 1| + |x + 5| >= 6.

Case 2: Assume that -5 =< x < 1. In this case x - 1 < 0, and x + 5 >= 0.

So |x - 1| = -(x - 1) = 1 - x, and |x + 5| = x + 5. Then

|x - 1| + |x + 5| = 1 - x + x + 5 = 6 >= 6.

Case 3: Assume that x >= 1. In this case x - 1 < 0, and x + 5 >= 0. So

|x - 1| = x -1 and |x + 5| = x + 5. Then

|x - 1| + |x+5| = x - 1 + x + 5 = 2x + 4 >= 2(1) + 4 = 6

So in all cases |x - 1| + |x + 5| >= 6.

QED

Proofs by Case (1)

Write clear and complete proofs by case for the following statement.

(1) If x is a real number, then |x + 3| - x > 2.

Proof. Assume that x is a real number.

- Case 1: Assume that x >= -3, so that x + 3 >= 0. In this case |x + 3| = x + 3. So we have

              |x + 3| - x = x + 3 - x = 3 > 2.

- Case 2: Now assume that x < - 3, so that x + 3 < 0. In this case |x + 3| = -(x + 3) = - 3 - x. So we have

|x + 3| - x = - 3 - x -x = - 2x - 3 > -2(-3) -3 = 6 - 3 = 3 > 2.

So in all cases |x+3| - x > 2.

QED