Equivalence of inequalities between means of opposite signs
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Suppose an average between power means with exponents p and q holds:
then:
We raise both sides to the power of -1 (strictly decreasing function in positive reals):
We get the inequality for means with exponents -p and -q, and we can use the same reasoning backwards, thus proving the inequalities to be equivalent, which will be used in some of the later proofs.
For any q the inequality between mean with exponent q and geometric mean can be transformed in the following way:
(the first inequality is to be proven for positive q, and the latter otherwise)
We raise both sides to the power of q:
in both cases we get the inequality between weighted arithmetic and geometric means for the sequence , which can be proved by Jensen's inequality, making use of the fact the logarithmic function is concave:
By applying (strictly increasing) exp function to both sides we get the inequality:
Thus for any postive q it is true that:
since the inequality holds for any q, however small, and, as will be shown later, the expressions on the left and right approximate the geometric mean better as q approaches 0, the limit of the power mean for q approaching 0 is the geometric mean:
Inequality between any two power means
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We are to prove that for any p<q the following inequality holds:
if p is negative, and q is positive, the inequality is equivalent to the one proved above:
The proof for positive p and q is as follows:
Define the following function: . f is a power function, so it does have a second derivative: which is strictly positive within the domain of f, since q > p, so we know f is convex.
Using this, and the Jensen's inequality we get:
after raising both side to the power of 1/q (an increasing function, since 1/q is positive) we get the inequality which was to be proven:
Using the previously shown equivalence we can prove the inequality for negative p and q by substituting them with, respectively, -q and -p, QED.
Minimum and maximum
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Minimum and maximum are assumed to be the power means with exponents of . Thus for any q:
For maximum the proof is as follows:
Assume WLoG that the sequence xi is nonincreasing and no weight is zero.
Then the inequality is equivalent to:
After raising both sides to the power of q we get (depending on the sign of q) one of the inequalities:
≤ for q>0, ≥ for q<0.
After substracting from the both sides we get:
After dividing by :
1 - w1 is nonzero, thus:
Substacting x1 leaves:
which is obvious, since x1 is greater or equal to any xi, and thus:
For minimum the proof is almost the same, only instead of x1, w1 we use xn, wn, QED.