It is called triangle inequality to the property of two real numbers that consist in that the absolute value of their sum is always less than or equal to the sum of their absolute values. This property is also known as Minkowski inequality or triangular inequality.
This property of numbers is called triangular inequality because in triangles it happens that the length of one side is always less than or equal to the sum of the other two, even though this inequality does not always apply in the area of triangles..
There are several proofs of the triangular inequality in real numbers, but in this case we will choose one based on the properties of absolute value and the binomial squared.
Theorem: For every pair of numbers to Y b pertaining to the real numbers it has to:
| a + b | ≤ | to | + | b |
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We begin by considering the first member of the inequality, which will be squared:
| a + b | ^ 2 = (a + b) ^ 2 = a ^ 2 + 2 a b + b ^ 2 (Eq. 1)
In the previous step we used the property that any number squared is equal to the absolute value of said number squared, that is: | x | ^ 2 = x ^ 2. The development of the squared binomial has also been used.
All number x is less than or equal to its absolute value. If the number is positive it is equal, but if the number is negative it will always be less than a positive number. In this case its own absolute value, that is to say that it can be stated that x ≤ | x |.
The product (a b) is a number, therefore it applies that (a b) ≤ | a b |. When this property is applied to (Eq. 1) we have:
| a + b | ^ 2 = a ^ 2 + 2 (a b) + b ^ 2 ≤ a ^ 2 + 2 | a b | + b ^ 2 (Eq. 2)
Taking into account that | a b | = | a || b | (Eq. 2) can be written as follows:
| a + b | ^ 2 ≤ a ^ 2 + 2 | a || b | + b ^ 2 (Eq. 3)
But since we said before that the square of a number is equal to the absolute value of the number squared, then equation 3 can be rewritten as follows:
| a + b | ^ 2 ≤ | a | ^ 2 + 2 | a | | b | + | b | ^ 2 (Eq. 4)
In the second member of the inequality, a remarkable product is recognized, which when applied leads to:
| a + b | ^ 2 ≤ (| a | + | b |) ^ 2 (Eq. 5)
In the previous expression, it should be noted that the values to be squared in both members of the inequality are positive, therefore it must also be satisfied that:
| a + b | ≤ (| a | + | b |) (Eq. 6)
The above expression is exactly what was wanted to demonstrate.
Next we will check the triangular inequality with several examples.
We take the value a = 2 and the value b = 5, that is, both positive numbers and we check whether the inequality is satisfied or not.
| 2 + 5 | ≤ | 2 | + | 5 |
| 7 | ≤ | 2 | + | 5 |
7 ≤ 2+ 5
Equality is verified, therefore the triangle inequality theorem has been fulfilled.
The following values are chosen a = 2 and b = -5, that is, a positive number and the other negative, we check whether the inequality is satisfied or not.
| 2 - 5 | ≤ | 2 | + | -5 |
| -3 | ≤ | 2 | + | -5 |
3 ≤ 2 + 5
The inequality is satisfied, therefore the triangular inequality theorem has been verified.
We take the value a = -2 and the value b = 5, that is, a negative number and the other positive, we check whether the inequality is satisfied or not.
| -2 + 5 | ≤ | -2 | + | 5 |
| 3 | ≤ | -2 | + | 5 |
3 ≤ 2 + 5
The inequality is verified, therefore the theorem has been fulfilled.
The following values a = -2 and b = -5 are chosen, that is, both negative numbers and we check whether the inequality is satisfied or not.
| -2 - 5 | ≤ | -2 | + | -5 |
| -7 | ≤ | -2 | + | -5 |
7 ≤ 2+ 5
Equality is verified, therefore Minkowski's inequality theorem has been fulfilled.
We take the value a = 0 and the value b = 5, that is, a number zero and the other positive, then we check whether the inequality is satisfied or not.
| 0 + 5 | ≤ | 0 | + | 5 |
| 5 | ≤ | 0 | + | 5 |
5 ≤ 0+ 5
Equality is fulfilled, therefore the triangle inequality theorem has been verified.
We take the value a = 0 and the value b = -7, that is, a number zero and the other positive, then we check whether the inequality is satisfied or not.
| 0 - 7 | ≤ | 0 | + | -7 |
| -7 | ≤ | 0 | + | -7 |
7 ≤ 0+ 7
Equality is verified, therefore the triangular inequality theorem has been fulfilled.
In the following exercises, represent geometrically the triangle inequality or Minkowski inequality for the numbers a and b.
The number a will be represented as a segment on the X axis, its origin O coincides with the zero of the X axis and the other end of the segment (at point P) will be in the positive direction (to the right) of the X axis if a > 0, but if a < 0 estará hacia la dirección negativa del eje X, tantas unidades como indique su valor absoluto.
Similarly, the number b will be represented as a segment whose origin is on point P. The other extreme, that is, point Q will be to the right of P if b is positive (b> 0) and point Q will be | b | units to the left of P if b<0.
Graph the triangle inequality for a = 5 and b = 3 | a + b | ≤ | to | + | b |, being c = a + b.
Graph the triangular inequality for a = 5 and b = -3.
| a + b | ≤ | to | + | b |, being c = a + b.
Graphically show the triangle inequality for a = -5 and b = 3.
| a + b | ≤ | to | + | b |, being c = a + b.
Graphically construct the triangular inequality for a = -5 and b = -3.
| a + b | ≤ | to | + | b |, being c = a + b.
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