prove that geometric mean is Greater or equal to harmonic mean
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Step-by-step explanation:
In mathematics, the inequality of arithmetic and geometric means, or more briefly the AM–GM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list; and further, that the two means are equal if and only if every number in the list is the same.
The simplest non-trivial case — i.e., with more than one variable — for two non-negative numbers x and y, is the statement that
{\displaystyle {\frac {x+y}{2}}\geq {\sqrt {xy}}}{\frac {x+y}2}\geq {\sqrt {xy}}
with equality if and only if x = y. This case can be seen from the fact that the square of a real number is always non-negative (greater than or equal to zero) and from the elementary case (a ± b)2 = a2 ± 2ab + b2 of the binomial formula:
{\displaystyle {\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}}{\begin{aligned}0&\leq (x-y)^{2}\\&=x^{2}-2xy+y^{2}\\&=x^{2}+2xy+y^{2}-4xy\\&=(x+y)^{2}-4xy.\end{aligned}}
Hence (x + y)2 ≥ 4xy, with equality precisely when (x − y)2 = 0, i.e. x = y. The AM-GM inequality then follows from taking the positive square root of both sides.
For a geometrical interpretation, consider a rectangle with sides of length x and y, hence it has perimeter 2x + 2y and area xy. Similarly, a square with all sides of length √xy has the perimeter 4√xy and the same area as the rectangle. The simplest non-trivial case of the AM–GM inequality implies for the perimeters that 2x + 2y ≥ 4√xy and that only the square has the smallest perimeter amongst all rectangles of equal area.
Extensions of the AM–GM inequality are available to include weights or generalized means.
Given any collection of real numbers, an average is a single number intended to give an estimate of the general magnitude of the numbers. Formally, it is a function from a set of n numbers to a single number with the following properties:
If all the numbers are equal, their average should also equal this value: AV(x, x, x, ...) = x.
The average must not exceed the maximum of the numbers nor be less than their minimum. We may wish to be stricter and say that if not all the numbers are equal, it must be greater than the minimum and less than the maximum, but this would rule out the median as an average, since the median of say (1,1,1,1,2) is 1.
The average must be multiplicatively linear, i.e. if all numbers are multiplied by the same constant k their average will be multiplied by the same number: AV(kx, ky) = k.AV(x, y).
The average must be order invariant; if we permute the numbers, it will not change their average: AV(y, x) = AV(x, y). This rules out just picking the first number, or the average of the first and last, or other weighted averages.
The average must be monotonic; if any one number increases (the others being unchanged), the average must not decrease, and vice versa. This rules out some "robust measures", where outliers are rejected before the average is taken. We may wish to be stricter and say that if any number increases, so must the average. Again, this would rule out the median as an average, since the median of say (1,1,1,1,2) and (1,1,1,2,2) are both 1.
It might be supposed that the average should be translation invariant, so that if all numbers are increased by the same constant k their average will increase by the same number: AV(x+k,y+k) = AV(x,y)+k. However, it can be shown that there is only one average meeting this and the other requirements strictly: the arithmetic mean discussed below. If the less strict versions of the requirements are used, the median and other quantiles would meet all the requirements.
It might also be supposed that the average should be a continuous function of the numbers. Again, this would rule out quantiles.
Hope it helps