# proof of arithmetic-geometric means inequality using Lagrange multipliers

As an interesting example of the Lagrange multiplier method, we employ it to prove the arithmetic-geometric means inequality:

$$\sqrt[n]{{x}_{1}\mathrm{\cdots}{x}_{n}}\le \frac{{x}_{1}+\mathrm{\cdots}+{x}_{n}}{n},{x}_{i}\ge 0,$$ |

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

To begin with, define $f:{\mathbb{R}}_{\ge 0}^{n}\to {\mathbb{R}}_{\ge 0}$ by $f(x)={({x}_{1}\mathrm{\cdots}{x}_{n})}^{1/n}$.

Fix $a>0$ (the arithmetic mean^{}),
and consider the set $M=\{x\in {\mathbb{R}}^{n}:{x}_{i}\ge 0\text{for all}i\text{,}{\sum}_{i}{x}_{i}=na\}$.
This is clearly a closed and bounded set in ${\mathbb{R}}^{n}$, and hence is compact^{}.
Therefore $f$ on $M$ has both a maximum and minimum.

$M$ is almost a manifold (a differentiable^{} surface), except that it has a boundary consisting
of points with ${x}_{i}=0$ for some $i$.
Clearly $f$ attains the minimum zero on this boundary,
and on the “interior” of $M$,
that is ${M}^{\prime}=\{x\in {\mathbb{R}}^{n}:{x}_{i}>0\text{for all}i\text{,}{\sum}_{i}{x}_{i}=na\}$,
$f$ is positive^{}. Now ${M}^{\prime}$ is a manifold, and the maximum of $f$ on $M$
is evidently a local maximum^{} on ${M}^{\prime}$. Hence
the Lagrange multiplier method may be applied.

We have the constraint^{1}^{1}Although $g$ does not constrain that ${x}_{i}>0$,
this does not really matter — for those who are worried about such
things, a rigorous way to escape this annoyance is to simply
define $f(x)=0$ to be zero whenever ${x}_{i}\le 0$ for some $i$.
Clearly the new $f$ cannot have a maximum at such points,
so we may simply ignore the points with ${x}_{i}\le 0$ when solving the system.
And for the same reason, the fact that $f$ fails to be differentiable at the boundary
points is immaterial. $0=g(x)={\sum}_{i}{x}_{i}-na$
under which we are to maximize $f$.

We compute:

$\frac{f(x)}{n{x}_{i}}}={\displaystyle \frac{\partial f}{\partial {x}_{i}}}=\lambda {\displaystyle \frac{\partial g}{\partial {x}_{i}}}=\lambda .$ |

If we take the product^{}, $i=1,\mathrm{\dots},n$, of the extreme left- and right- hand sides of this equation,
we obtain

$$\frac{1}{{n}^{n}}=\frac{f{(x)}^{n}}{{n}^{n}{x}_{1}\mathrm{\cdots}{x}_{n}}={\lambda}^{n}$$ |

from which it follows that $\lambda =1/n$ ($\lambda =-1/n$ is obviously inadmissible). Then substituting back, we get $f(x)={x}_{i}$, which implies that the ${x}_{i}$ are all equal to $a$, The value of $f$ at this point is $a$.

Since we have obtained a unique solution to the Lagrange multiplier system, this unique solution must be the unique maximum point of $f$ on $M$. So $f(x)\le a$ amongst all numbers ${x}_{i}$ whose arithmetic mean is $a$, with equality occurring if and only if all the ${x}_{i}$ are equal to $a$.

## Proof of concavity

The question of whether the point obtained from solving Lagrange system
is actually a maximum for $f$ was taken care of by our preliminary compactness argument^{}. Another popular way to determine whether a given stationary point
is a local maximum or minimum is by studying the Hessian^{} of the function
at that point.
For the sake of illustrating the relevant techniques,
we will prove again that ${x}_{i}=a$ is a local maximum for $f$,
this time by showing that the Hessian is negative definite^{}.

Actually, it turns out to be not much
harder to show that $f$ is weakly concave^{} *everywhere*
on ${\mathbb{R}}_{\ge 0}^{n}$, not just on $M$.
A plausibility argument for this fact
is that the graph of $t\mapsto \sqrt[n]{t}$
is concave, and since $f$ also involves an $n$th root,
it should be concave too. We will prove concavity of $f$ by
showing that the Hessian of $f$ is negative semi-definite.

Since $M$ is a *convex*^{2}^{2}
If $M$ is not convex, then the arguments that follow will not work.
In general, a prescription to determine whether a critical point
is a local minimum or maximum can be found
in tests for local extrema in Lagrange multiplier method. set, the restriction^{}
of $f$ to $M$ will also be weakly concave;
then we can conclude that
the critical point ${x}_{i}=a$ on $M$
must be a *global* minimum on $M$.

We begin by calculating second-order derivatives^{}:

$\frac{{\partial}^{2}f}{\partial {x}_{j}\partial {x}_{i}}$ | $={\displaystyle \frac{\partial}{\partial {x}_{j}}}\left({\displaystyle \frac{f(x)}{n{x}_{i}}}\right)={\displaystyle \frac{1}{n{x}_{i}}}{\displaystyle \frac{\partial f}{\partial {x}_{j}}}+f(x){\displaystyle \frac{\partial}{\partial {x}_{j}}}{\displaystyle \frac{1}{n{x}_{i}}}$ | ||

$={\displaystyle \frac{f(x)}{{n}^{2}{x}_{i}{x}_{j}}}-{\delta}_{ij}{\displaystyle \frac{f(x)}{n{x}_{i}^{2}}}.$ |

(The symbol ${\delta}_{ij}$ is the Kronecker delta.) Thus the Hessian matrix is:

${\mathrm{D}}^{2}f(x)={\displaystyle \frac{f(x)}{{n}^{2}}}{\left[{\displaystyle \frac{1}{{x}_{i}{x}_{j}}}\right]}_{ij}-{\displaystyle \frac{f(x)}{{n}^{2}}}\left[\begin{array}{c}\hfill \frac{n}{{x}_{1}^{2}}\hfill \\ \hfill \hfill & \hfill \mathrm{\ddots}\hfill \\ \hfill \hfill & \hfill \hfill & \hfill \frac{n}{{x}_{n}^{2}}\hfill \end{array}\right]$ |

Now if we had actual numbers to substitute for the ${x}_{i}$,
then we can go to a computer and ask if the matrix is negative definite or not.
But we want to prove global concavity, so we have to employ some clever
algebra^{}, and work directly from the definition.

Let $v=({v}_{1},\mathrm{\dots},{v}_{n})$ be a test vector. Negative semi-definiteness means $v{\mathrm{D}}^{2}f(x){v}^{\mathrm{T}}\le 0$ for all $v$. So let us write out:

$v{\mathrm{D}}^{2}f(x){v}^{\mathrm{T}}$ | $={\displaystyle \frac{f(x)}{{n}^{2}}}\left({\displaystyle \sum _{i,j}}{\displaystyle \frac{{v}_{i}}{{x}_{i}}}{\displaystyle \frac{{v}_{j}}{{x}_{j}}}-n{\displaystyle \sum _{i}}{\displaystyle \frac{{v}_{i}^{2}}{{x}_{i}^{2}}}\right)$ | ||

$={\displaystyle \frac{f(x)}{{n}^{2}}}\left({\left({\displaystyle \sum _{i}}{\displaystyle \frac{{v}_{i}}{{x}_{i}}}\right)}^{2}-n{\displaystyle \sum _{i}}{\left({\displaystyle \frac{{v}_{i}}{{x}_{i}}}\right)}^{2}\right).$ |

We would be done if we can prove

$${\left(\sum _{i}{w}_{i}\right)}^{2}\le n\sum _{i}{w}_{i}^{2},{w}_{i}=\frac{{v}_{i}}{{x}_{i}}.$$ | (1) |

But look, this is just the Cauchy-Schwarz inequality,
concerning the dot product^{} of the vectors $({w}_{1},\mathrm{\dots},{w}_{n})$
and $(1,\mathrm{\dots},1)$!
So ${\mathrm{D}}^{2}f(x)$ is negative semi-definite everywhere,
i.e. $f$ is weakly concave.

Moreover, the case of equality in the Cauchy-Schwarz inequality (1)
only occurs when $({w}_{1},\mathrm{\dots},{w}_{n})$ is parallel^{} to $(1,\mathrm{\dots},1)$,
i.e. ${v}_{i}=\mu {x}_{i}$ for some scalar $\mu $.
Notice that $(1,\mathrm{\dots},1)$ is the normal vector to
the tangent space^{} of $M$,
so $(1,\mathrm{\dots},1)\cdot v=\mu {\sum}_{i}{x}_{i}\ne 0$
unless $\mu =0$.
This means, equality never holds in (1)
if $v\ne 0$ is *restricted to the tangent space* of $M$.
In turn, this means that $f$ is *strictly* concave
on the convex set $M$, and
so ${x}_{i}=a$ is a strict global minimum of $f$ on $M$.

## Proof of inequality by symmetry argument

Observe that the maximum point ${x}_{i}=a$ is pleasantly symmetric^{}.
We can show that the solution has to be symmetric,
by exploiting the symmetry^{} of the function $f$ and the set $M$.

We have already calculated that $f$ is strictly concave on $M$; this was independent of the Lagrange multiplier method. Since $f$ is strictly concave on a closed convex set, it has a unique maximum there; call it $x$.

Pick any indices $i$ and $j$, ranging from $1$ to $n$, and form
the linear transformation $T$, which switches the $i$th and $j$th coordinates^{}
of its argument.

By definition, $f(x)\ge f(\xi )$ for all $\xi \in M$.
But $f\circ T=f$, so this means $f(Tx)\ge f(T\xi )$ for all $\xi \in M$.
But $M$ is mapped to itself by the transformation^{} $T$,
so this is the same as saying that $f(Tx)\ge f(\xi )$ for all $\xi \in M$.
But the global maximum is unique, so $Tx=x$,
that is, ${x}_{i}={x}_{j}$.

Note that the argument in this last proof
is extremely general — it takes the form:
if $x$ is the unique maximum of $f:M\to \mathbb{R}$,
and $T$ is any symmetry of $M$ such that $f\circ T=f$,
then $Tx=x$. This kind of symmetry argument was used to great effect
by Jakob Steiner in his famous attempted proof of the isoperimetric inequality^{}:
among all closed curves of a given arc length^{}, the circle maximizes
the area enclosed. However, Steiner did not realize that
the assumption^{} that there is a unique maximizing curve
has to be proved. Actually, that is the hardest part of the proof —
if the assumption is known, then one may easily calculate using the *Lagrange multiplier method*
that the maximizing curve must be a circle!

Title | proof of arithmetic-geometric means inequality using Lagrange multipliers |
---|---|

Canonical name | ProofOfArithmeticgeometricMeansInequalityUsingLagrangeMultipliers |

Date of creation | 2013-03-22 15:25:53 |

Last modified on | 2013-03-22 15:25:53 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 16 |

Author | stevecheng (10074) |

Entry type | Example |

Classification | msc 49-00 |

Classification | msc 26B12 |

Related topic | ArithmeticGeometricMeansInequality |

Related topic | UsingJensensInequalityToProveTheArithmeticGeometricHarmonicMeansInequality |