User:Ghazer~enwiki/avgs

Suppose an average between power means with exponents p and q holds:

${\displaystyle {\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

then:

${\displaystyle {\sqrt[{p}]{\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{p}}}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}{\frac {w_{i}}{x_{i}^{q}}}}}}$

We raise both sides to the power of -1 (strictly decreasing function in positive reals):

${\displaystyle {\sqrt[{-p}]{\sum _{i=1}^{n}w_{i}x_{i}^{-p}}}={\sqrt[{p}]{\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{p}}}}}}\geq {\sqrt[{q}]{\frac {1}{\sum _{i=1}^{n}w_{i}{\frac {1}{x_{i}^{q}}}}}}={\sqrt[{-q}]{\sum _{i=1}^{n}w_{i}x_{i}^{-q}}}}$

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.

For any q the inequality between mean with exponent q and geometric mean can be transformed in the following way:

${\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

${\displaystyle {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}}$

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

We raise both sides to the power of q:

${\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}\cdot q}\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}}$

${\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}\cdot q}\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}}$

in both cases we get the inequality between weighted arithmetic and geometric means for the sequence ${\displaystyle x_{i}^{q}}$, which can be proved by Jensen's inequality, making use of the fact the logarithmic function is concave:

${\displaystyle \sum _{i=1}^{n}w_{i}\log(x_{i})\leq \log(\sum _{i=1}^{n}w_{i}x_{i})}$

${\displaystyle log(\prod _{i=1}^{n}x_{i}^{w_{i}})\leq log(\sum _{i=1}^{n}w_{i}x_{i})}$

By applying (strictly increasing) exp function to both sides we get the inequality:

${\displaystyle \prod _{i=1}^{n}x_{i}^{w_{i}}\leq \sum _{i=1}^{n}w_{i}x_{i}}$

Thus for any postive q it is true that:

${\displaystyle {\sqrt[{-q}]{\sum _{i=1}^{n}w_{i}x_{i}^{-q}}}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

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:

${\displaystyle \lim _{q\rightarrow 0}{\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}=\prod _{i=1}^{n}x_{i}^{w_{i}}}$

We are to prove that for any p<q the following inequality holds:

${\displaystyle {\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

if p is negative, and q is positive, the inequality is equivalent to the one proved above:

${\displaystyle {\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq \prod _{i=1}^{n}x_{i}^{w_{i}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

The proof for positive p and q is as follows: Define the following function: ${\displaystyle f:{\mathbb {R} _{+}}\rightarrow {\mathbb {R} _{+}},}$ ${\displaystyle f(x)=x^{\frac {q}{p}}}$. f is a power function, so it does have a second derivative: ${\displaystyle f''(x)=({\frac {q}{p}})({\frac {q}{p}}-1)x^{{\frac {q}{p}}-2},}$ 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:

${\displaystyle f(\sum _{i=1}^{n}w_{i}x_{i}^{p})\leq \sum _{i=1}^{n}w_{i}f(x_{i}^{p})}$

${\displaystyle {\sqrt[{\frac {p}{q}}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq \sum _{i=1}^{n}w_{i}x_{i}^{q}}$

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:

${\displaystyle {\sqrt[{p}]{\sum _{i=1}^{n}w_{i}x_{i}^{p}}}\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}}$

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 are assumed to be the power means with exponents of ${\displaystyle -/+\infty }$. Thus for any q:

${\displaystyle \min(x_{1},x_{2},\ldots ,x_{n})\leq {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}\leq \max(x_{1},x_{2},\ldots ,x_{n})}$

For maximum the proof is as follows: Assume WLoG that the sequence x_i is nonincreasing and no weight is zero.

Then the inequality is equivalent to:

${\displaystyle {\sqrt[{q}]{\sum _{i=1}^{n}w_{i}x_{i}^{q}}}\leq x_{1}}$

After raising both sides to the power of q we get (depending on the sign of q) one of the inequalities:

${\displaystyle \sum _{i=1}^{n}w_{i}x_{i}^{q}\leq {\color {red}\geq }x_{1}^{q}}$

≤ for q>0, ≥ for q<0.

After substracting ${\displaystyle w_{1}x_{1}}$ from the both sides we get:

${\displaystyle \sum _{i=2}^{n}w_{i}x_{i}^{q}\leq {\color {red}\geq }(1-w_{1})x_{1}^{q}}$

After dividing by ${\displaystyle (1-w_{1})}$:

${\displaystyle \sum _{i=2}^{n}{\frac {w_{i}}{(1-w_{1})}}x_{i}^{q}\leq {\color {red}\geq }x_{1}^{q}}$

1 - w₁ is nonzero, thus:

${\displaystyle \sum _{i=2}^{n}{\frac {w_{i}}{(1-w_{1})}}=1}$

Substacting x₁ leaves:

${\displaystyle \sum _{i=2}^{n}{\frac {w_{i}}{(1-w_{1})}}(x_{i}^{q}-x_{1}^{q})\leq {\color {red}\geq }0}$

which is obvious, since x₁ is greater or equal to any x_i, and thus:

${\displaystyle x_{i}^{q}-x_{1}^{q}\leq {\color {red}\geq }0}$

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