The quadratic mean of two real numbers x and y equals p (x 2 y 2) / 2. By computing the arithmetic and quadratic means of different pairs of positive real numbers, formulate a conjecture about their relative sizes and prove your conjecture.?

The quadratic mean of 2 real positive numbers is greater than or equal to the arithmetic mean.

Step-by-step explanation:

x and y Quadratic Mean Arithmetic mean

3 and 3 3 3

2 and 3 2.55 2.5

3 and 6 4.74 4.5

2 and 5 3.8 3.5

2 and 17 12.1 9.5

18 and 28 23.5 23

10 and 48 34.7 29

The quadratic mean is always greater than the arithmetic mean except when x and y are the same.

When the difference between the pairs is small the difference in the means is also small. As that difference increases the difference in the means also increases.

So we conjecture that the quadratic mean is always greater than or equal to the arithmetic mean.

Proof.

Suppose it is true then:

√ (x^2 + y^2) / 2) ≥ (x + y) / 2 Squaring both sides:

(x ^2 + y^2) / 2 ≥ (x + y) ^2 / 4 Multiply through by 4:

2x^2 + 2y^2 ≥ (x + y) ^2

2x^2 + 2y^2 > = x^2 + 2xy + y^2

x^2 + y^2 > = 2xy.

x^2 - 2xy + y^2 ≥ 0

(x - y) ^2 ≥ 0

This is true because the square of any real number is positive so the original inequality must also be true.