# proof of general means inequality

Let ${w}_{1}$, ${w}_{2}$, …, ${w}_{n}$ be positive real numbers such that ${w}_{1}+{w}_{2}+\mathrm{\cdots}+{w}_{n}=1$. For any real number $r\ne 0$, the weighted power mean of degree $r$ of $n$ positive real numbers ${x}_{1}$, ${x}_{2}$, …, ${x}_{n}$ (with respect to the weights ${w}_{1}$, …, ${w}_{n}$) is defined as

$${M}_{w}^{r}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})={({w}_{1}{x}_{1}^{r}+{w}_{2}{x}_{2}^{r}+\mathrm{\cdots}+{w}_{n}{x}_{n}^{r})}^{1/r}.$$ |

The definition is extended to the case $r=0$ by taking the limit $r\to 0$; this yields the weighted geometric mean

$${M}_{w}^{0}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})={x}_{1}^{{w}_{1}}{x}_{2}^{{w}_{2}}\mathrm{\dots}{x}_{n}^{{w}_{n}}$$ |

(see derivation of zeroth weighted power mean). We will prove the
weighted power means inequality^{}, which states that for any two real
numbers $$, the weighted power means of orders $r$ and $s$
of $n$ positive real numbers ${x}_{1}$, ${x}_{2}$, …, ${x}_{n}$ satisfy the
inequality

$${M}_{w}^{r}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})\le {M}_{w}^{s}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})$$ |

with equality if and only if all the ${x}_{i}$ are equal.

First, let us suppose that $r$ and $s$ are nonzero. We distinguish three cases for the signs of $r$ and $s$: $$, $$, and $$. Let us consider the last case, i.e. assume $r$ and $s$ are both positive; the others are similar. We write $t=\frac{s}{r}$ and ${y}_{i}={x}_{i}^{r}$ for $1\le i\le n$; this implies ${y}_{i}^{t}={x}_{i}^{s}$. Consider the function

$f:(0,\mathrm{\infty})$ | $\to $ | $(0,\mathrm{\infty})$ | ||

$x$ | $\mapsto $ | ${x}^{t}.$ |

Since $t>1$, the second derivative of $f$ satisfies ${f}^{\prime \prime}(x)=t(t-1){x}^{t-2}>0$ for all $x>0$, so $f$ is a strictly convex function. Therefore, according to Jensen’s inequality,

${({w}_{1}{y}_{1}+{w}_{2}{y}_{2}+\mathrm{\cdots}+{w}_{n}{y}_{n})}^{t}$ | $=$ | $f({w}_{1}{y}_{1}+{w}_{2}{y}_{2}+\mathrm{\cdots}+{w}_{n}{y}_{n})$ | ||

$\le $ | ${w}_{1}f({y}_{1})+{w}_{2}f({y}_{2})+\mathrm{\cdots}+{w}_{n}f({y}_{n})$ | |||

$=$ | ${w}_{1}{y}_{1}^{t}+{w}_{2}{y}_{2}^{t}+\mathrm{\cdots}+{w}_{n}{y}_{n}^{t},$ |

with equality if and only if ${y}_{1}={y}_{2}=\mathrm{\cdots}={y}_{n}$. By substituting $t=\frac{s}{r}$ and ${y}_{i}={x}_{i}^{r}$ back into this inequality, we get

$${({w}_{1}{x}_{1}^{r}+{w}_{2}{x}_{2}^{r}+\mathrm{\cdots}+{w}_{n}{x}_{n}^{r})}^{s/r}\le {w}_{1}{x}_{1}^{s}+{w}_{2}{x}_{2}^{s}+\mathrm{\cdots}+{w}_{n}{x}_{n}^{s}$$ |

with equality if and only if ${x}_{1}={x}_{2}=\mathrm{\cdots}={x}_{n}$. Since $s$ is positive, the function $x\mapsto {x}^{1/s}$ is strictly increasing, so raising both sides to the power $1/s$ preserves the inequality:

$${({w}_{1}{x}_{1}^{r}+{w}_{2}{x}_{2}^{r}+\mathrm{\cdots}+{w}_{n}{x}_{n}^{r})}^{1/r}\le {({w}_{1}{x}_{1}^{s}+{w}_{2}{x}_{2}^{s}+\mathrm{\cdots}+{w}_{n}{x}_{n}^{s})}^{1/s},$$ |

which is the inequality we had to prove. Equality holds if and only if all the ${x}_{i}$ are equal.

If $r=0$, the inequality is still correct: ${M}_{w}^{0}$ is defined as
${lim}_{r\to 0}{M}_{w}^{r}$, and since ${M}_{w}^{r}\le {M}_{w}^{s}$ for all $$ with
$r\ne 0$, the same holds for the limit $r\to 0$. The same argument^{}
shows that the inequality also holds for $s=0$, i.e. that
${M}_{w}^{r}\le {M}_{w}^{0}$ for all $$. We conclude that for all real numbers
$r$ and $s$ such that $$,

$${M}_{w}^{r}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n})\le {M}_{w}^{s}({x}_{1},{x}_{2},\mathrm{\dots},{x}_{n}).$$ |

Title | proof of general means inequality |

Canonical name | ProofOfGeneralMeansInequality |

Date of creation | 2013-03-22 13:10:26 |

Last modified on | 2013-03-22 13:10:26 |

Owner | pbruin (1001) |

Last modified by | pbruin (1001) |

Numerical id | 5 |

Author | pbruin (1001) |

Entry type | Proof |

Classification | msc 26D15 |

Related topic | ArithmeticMean |

Related topic | GeometricMean |

Related topic | HarmonicMean |

Related topic | RootMeanSquare3 |

Related topic | PowerMean |

Related topic | WeightedPowerMean |

Related topic | ArithmeticGeometricMeansInequality |

Related topic | JensensInequality |