Editing Generalized mean (section)

===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.