User:Ghazer~enwiki/avgs

Equivalence of inequalities between means of opposite signs
Suppose an average between power means with exponents p and q holds:
 * $$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$

then:
 * $$\sqrt[p]{\sum_{i=1}^n\frac{w_i}{x_i^p}}\leq \sqrt[q]{\sum_{i=1}^n\frac{w_i}{x_i^q}}$$

We raise both sides to the power of -1 (strictly decreasing function in positive reals):
 * $$\sqrt[-p]{\sum_{i=1}^nw_ix_i^{-p}}=\sqrt[p]{\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^p}}}\geq \sqrt[q]{\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^q}}}=\sqrt[-q]{\sum_{i=1}^nw_ix_i^{-q}}$$

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.

Geometric mean
For any q the inequality between mean with exponent q and geometric mean can be transformed in the following way:
 * $$\prod_{i=1}^nx_i^{w_i} \leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$
 * $$\sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq \prod_{i=1}^nx_i^{w_i} $$

(the first inequality is to be proven for positive q, and the latter otherwise)

We raise both sides to the power of q:
 * $$\prod_{i=1}^nx_i^{w_i\cdot q} \leq \sum_{i=1}^nw_ix_i^q$$
 * $$\prod_{i=1}^nx_i^{w_i\cdot q} \leq \sum_{i=1}^nw_ix_i^q$$

in both cases we get the inequality between weighted arithmetic and geometric means for the sequence $$x_i^q$$, which can be proved by Jensen's inequality, making use of the fact the logarithmic function is concave:
 * $$\sum_{i=1}^nw_i\log(x_i) \leq \log(\sum_{i=1}^nw_ix_i)$$
 * $$log(\prod_{i=1}^nx_i^{w_i}) \leq log(\sum_{i=1}^nw_ix_i)$$

By applying (strictly increasing) exp function to both sides we get the inequality:
 * $$\prod_{i=1}^nx_i^{w_i} \leq \sum_{i=1}^nw_ix_i$$

Thus for any postive q it is true that:
 * $$\sqrt[-q]{\sum_{i=1}^nw_ix_i^{-q}}\leq \prod_{i=1}^nx_i^{w_i} \leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$

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:
 * $$\lim_{q\rightarrow 0}\sqrt[q]{\sum_{i=1}^nw_ix_i^{q}}=\prod_{i=1}^nx_i^{w_i}$$

Inequality between any two power means
We are to prove that for any p<q the following inequality holds:
 * $$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$

if p is negative, and q is positive, the inequality is equivalent to the one proved above:
 * $$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq \prod_{i=1}^nx_i^{w_i} \leq\sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$

The proof for positive p and q is as follows: Define the following function: $$f:{\mathbb R_+}\rightarrow{\mathbb R_+},$$ $$f(x)=x^{\frac{q}{p}}$$. f is a power function, so it does have a second derivative: $$f(x)=(\frac{q}{p})(\frac{q}{p}-1)x^{\frac{q}{p}-2},$$ 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:
 * $$f(\sum_{i=1}^nw_ix_i^p)\leq\sum_{i=1}^nw_if(x_i^p)$$
 * $$\sqrt[\frac{p}{q}]{\sum_{i=1}^nw_ix_i^p}\leq\sum_{i=1}^nw_ix_i^q$$

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:
 * $$\sqrt[p]{\sum_{i=1}^nw_ix_i^p}\leq\sqrt[q]{\sum_{i=1}^nw_ix_i^q}$$

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
Minimum and maximum are assumed to be the power means with exponents of $$-/+\infty$$. Thus for any q:
 * $$\min (x_1,x_2,\ldots ,x_n)\leq \sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq \max (x_1,x_2,\ldots ,x_n)$$

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:
 * $$\sqrt[q]{\sum_{i=1}^nw_ix_i^q}\leq x_1$$

After raising both sides to the power of q we get (depending on the sign of q) one of the inequalities:
 * $$\sum_{i=1}^nw_ix_i^q\leq {\color{red} \geq} x_1^q$$

≤ for q>0, ≥ for q<0.

After substracting $$w_1x_1$$ from the both sides we get:
 * $$\sum_{i=2}^nw_ix_i^q\leq {\color{red} \geq} (1-w_1)x_1^q$$

After dividing by $$(1-w_1)$$:
 * $$\sum_{i=2}^n\frac{w_i}{(1-w_1)}x_i^q\leq {\color{red} \geq} x_1^q$$

1 - w1 is nonzero, thus:
 * $$\sum_{i=2}^n\frac{w_i}{(1-w_1)}=1$$

Substacting x1 leaves:
 * $$\sum_{i=2}^n\frac{w_i}{(1-w_1)}(x_i^q-x_1^q)\leq {\color{red} \geq} 0$$

which is obvious, since x1 is greater or equal to any xi, and thus:
 * $$x_i^q-x_1^q\leq {\color{red} \geq} 0$$

For minimum the proof is almost the same, only instead of x1, w1 we use xn, wn, QED.