Respuesta :

Answer:

The given statement is true.        

Step-by-step explanation:

We are given the following information in the question:

The average of three real numbers is greater than or equal to at least one of the numbers.

We can prove this statement.

Let a, b and c be the three real numbers.

Then, the average of these three real numbers is given by

[tex]\text{Average} = \displaystyle\frac{\text{Sum of all observation}}{\text{Total number of observation}}\\\\A = \frac{a+b+c}{3}\\\\3A = a + b + c[/tex]

Let a be the smallest of the three natural number.

Then, we can write,

[tex]a + b+c \geq a+a+a\\3A \geq 3a\\A\geq a[/tex]

Let a be the largest number.

[tex]a + b+c \leq a+a+a\\3A \leq 3a\\A\leq a[/tex]

Thus, looking at both the inequalities, we can say that the average of three real numbers is greater than or less than equal to one of the numbers.

It is proved that the average of three real numbers is greater than equal to at least one of the numbers .

What is the inequality equation?

Inequality equation is the equation in which the two expressions are compared with greater than, less than or other inequality signs.

Let check if average of three real numbers is greater than or equal to at least one of the numbers or not.

Let, [tex]x_1,x_2,x_3[/tex] are three real numbers and y is the average of these three number such as,

[tex]y=\dfrac{x_1+x_2+x_3}{3}[/tex]

Now let, y is less then all the three numbers as,

[tex]y < x_1, \\y < x_2,\\y < x_3[/tex]

Put this value in the above equation with less than sine pointing towards the average as,

[tex]y < \dfrac{y+y+y}{3}\\y < \dfrac{3y}{3}\\y < y[/tex]

This is a contradiction condition, which proves that the average of three real number is greater than or equal to at least one of the numer.

Thus, It is proved that the average of three real numbers is greater than equal to at least one of the numbers .

Learn more about the inequality equation here:

https://brainly.com/question/17724536

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