Editing Generalized mean (section)

==Proof of the weighted inequality==
We will prove the weighted power mean inequality. For the purpose of the proof we will assume the following without loss of generality:
<math display=block>\begin{align}
  w_i \in [0, 1] \\
  \sum_{i=1}^nw_i = 1
\end{align}</math>

The proof for unweighted power means can be easily obtained by substituting {{math|1= ''w<sub>i</sub>'' = 1/''n''}}.

===Equivalence of inequalities between means of opposite signs===
Suppose an average between power means with exponents {{mvar|p}} and {{mvar|q}} holds:
<math display="block">\left(\sum_{i=1}^n w_i x_i^p\right)^{1/p} \geq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}</math>
applying this, then:
<math display="block">\left(\sum_{i=1}^n\frac{w_i}{x_i^p}\right)^{1/p} \geq \left(\sum_{i=1}^n\frac{w_i}{x_i^q}\right)^{1/q}</math>

We raise both sides to the power of −1 (strictly decreasing function in positive reals):
<math display="block">\left(\sum_{i=1}^nw_ix_i^{-p}\right)^{-1/p}
= \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^p}}\right)^{1/p}
\leq \left(\frac{1}{\sum_{i=1}^nw_i\frac{1}{x_i^q}}\right)^{1/q}
= \left(\sum_{i=1}^nw_ix_i^{-q}\right)^{-1/q}</math>

We get the inequality for means with exponents {{math|−''p''}} and {{math|−''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 {{math|''q'' > 0}} and non-negative weights summing to 1, the following inequality holds:
<math display="block">\left(\sum_{i=1}^n w_i x_i^{-q}\right)^{-1/q} \leq \prod_{i=1}^n x_i^{w_i} \leq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}.</math>

The proof follows from [[Jensen's inequality]], making use of the fact the [[logarithm]] is concave:
<math display=block>\log \prod_{i=1}^n x_i^{w_i} = \sum_{i=1}^n w_i\log x_i \leq \log \sum_{i=1}^n w_i x_i.</math>

By applying the [[exponential function]] to both sides and observing that as a strictly increasing function it preserves the sign of the inequality, we get
<math display=block>\prod_{i=1}^n x_i^{w_i} \leq \sum_{i=1}^n w_i x_i.</math>

Taking {{mvar|q}}-th powers of the {{mvar|x<sub>i</sub>}} yields
<math display=block>\begin{align}
&\prod_{i=1}^n x_i^{q{\cdot}w_i} \leq \sum_{i=1}^n w_i x_i^q \\
&\prod_{i=1}^n x_i^{w_i} \leq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}.\end{align}</math>

Thus, we are done for the inequality with positive {{mvar|q}}; the case for negatives is identical but for the swapped signs in the last step:

<math display=block>\prod_{i=1}^n x_i^{-q{\cdot}w_i} \leq \sum_{i=1}^n w_i x_i^{-q}.</math>

Of course, taking each side to the power of a negative number {{math|-1/''q''}} swaps the direction of the inequality.

<math display=block>\prod_{i=1}^n x_i^{w_i} \geq \left(\sum_{i=1}^n w_i x_i^{-q}\right)^{-1/q}.</math>

===Inequality between any two power means===
We are to prove that for any {{math|''p'' < ''q''}} the following inequality holds:
<math display="block">\left(\sum_{i=1}^n w_i x_i^p\right)^{1/p} \leq \left(\sum_{i=1}^nw_ix_i^q\right)^{1/q}</math>
if {{mvar|p}} is negative, and {{mvar|q}} is positive, the inequality is equivalent to the one proved above:
<math display="block">\left(\sum_{i=1}^nw_i x_i^p\right)^{1/p} \leq \prod_{i=1}^n x_i^{w_i} \leq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}</math>

The proof for positive {{mvar|p}} and {{mvar|q}} is as follows: Define the following function: {{math|''f'' : '''R'''<sub>+</sub> → '''R'''<sub>+</sub>}} <math>f(x)=x^{\frac{q}{p}}</math>. {{mvar|f}} is a power function, so it does have a second derivative:
<math display="block">f''(x) = \left(\frac{q}{p} \right) \left( \frac{q}{p}-1 \right)x^{\frac{q}{p}-2}</math>
which is strictly positive within the domain of {{mvar|f}}, since {{math|''q'' > ''p''}}, so we know {{mvar|f}} is convex.

Using this, and the Jensen's inequality we get:
<math display="block">\begin{align}
     f \left( \sum_{i=1}^nw_ix_i^p \right) &\leq \sum_{i=1}^nw_if(x_i^p) \\[3pt]
  \left(\sum_{i=1}^n w_i x_i^p\right)^{q/p} &\leq \sum_{i=1}^nw_ix_i^q
\end{align}</math>
after raising both side to the power of {{math|1/''q''}} (an increasing function, since {{math|1/''q''}} is positive) we get the inequality which was to be proven:

<math display="block">\left(\sum_{i=1}^n w_i x_i^p\right)^{1/p} \leq \left(\sum_{i=1}^n w_i x_i^q\right)^{1/q}</math>

Using the previously shown equivalence we can prove the inequality for negative {{mvar|p}} and {{mvar|q}} by replacing them with {{mvar|&minus;q}} and {{mvar|&minus;p}}, respectively.