# 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:

 $\sqrt[n]{x_{1}\cdots x_{n}}\leq\frac{x_{1}+\cdots+x_{n}}{n}\,,\quad x_{i}\geq 0\,,$

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

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

Fix $a>0$ (the arithmetic mean  ), and consider the set $M=\{x\in\mathbb{R}^{n}:x_{i}\geq 0\text{ for all i, }\sum_{i}x_{i}=na\}$. 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 $M^{\prime}=\{x\in\mathbb{R}^{n}:x_{i}>0\text{ for all i, }\sum_{i}x_{i}=na\}$, $f$ 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 constraint11Although $g$ 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 $x_{i}\leq 0$ for some $i$. Clearly the new $f$ cannot have a maximum at such points, so we may simply ignore the points with $x_{i}\leq 0$ 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:

 $\displaystyle\frac{f(x)}{nx_{i}}=\frac{\partial f}{\partial x_{i}}=\lambda\,% \frac{\partial g}{\partial x_{i}}=\lambda\,.$

If we take the product   , $i=1,\ldots,n$, of the extreme left- and right- hand sides of this equation, we obtain

 $\frac{1}{n^{n}}=\frac{f(x)^{n}}{n^{n}x_{1}\cdots x_{n}}=\lambda^{n}$

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 $f(x)\leq a$ 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 $\mathbb{R}_{\geq 0}^{n}$, 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 convex22 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 $M$.

 $\displaystyle\frac{\partial^{2}f}{\partial x_{j}\partial x_{i}}$ $\displaystyle=\frac{\partial}{\partial x_{j}}\left(\frac{f(x)}{nx_{i}}\right)=% \frac{1}{nx_{i}}\frac{\partial f}{\partial x_{j}}+f(x)\frac{\partial}{\partial x% _{j}}\frac{1}{nx_{i}}$ $\displaystyle=\frac{f(x)}{n^{2}x_{i}x_{j}}-\delta_{ij}\frac{f(x)}{nx_{i}^{2}}\,.$

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

 $\displaystyle\operatorname{D}^{2}f(x)=\frac{f(x)}{n^{2}}\left[\frac{1}{x_{i}x_% {j}}\right]_{ij}-\frac{f(x)}{n^{2}}\begin{bmatrix}\frac{n}{x_{1}^{2}}\\ &\ddots\\ &&\frac{n}{x_{n}^{2}}\end{bmatrix}$

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

 $\displaystyle v\operatorname{D}^{2}f(x)v^{\mathrm{T}}$ $\displaystyle=\frac{f(x)}{n^{2}}\left(\sum_{i,j}\frac{v_{i}}{x_{i}}\frac{v_{j}% }{x_{j}}-n\sum_{i}\frac{v_{i}^{2}}{x_{i}^{2}}\right)$ $\displaystyle=\frac{f(x)}{n^{2}}\left(\left(\sum_{i}\frac{v_{i}}{x_{i}}\right)% ^{2}-n\sum_{i}\left(\frac{v_{i}}{x_{i}}\right)^{2}\right)\,.$

We would be done if we can prove

 $\left(\sum_{i}w_{i}\right)^{2}\leq n\sum_{i}w_{i}^{2}\,,\quad w_{i}=\frac{v_{i% }}{x_{i}}\,.$ (1)

But look, this is just the Cauchy-Schwarz inequality, concerning the dot product  of the vectors $(w_{1},\ldots,w_{n})$ and $(1,\ldots,1)$! So $\operatorname{D}^{2}f(x)$ 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 $(w_{1},\ldots,w_{n})$ is parallel    to $(1,\ldots,1)$, i.e. $v_{i}=\mu x_{i}$ for some scalar $\mu$. Notice that $(1,\ldots,1)$ is the normal vector to the tangent space  of $M$, so $(1,\ldots,1)\cdot v=\mu\sum_{i}x_{i}\neq 0$ unless $\mu=0$. This means, equality never holds in (1) if $v\neq 0$ 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

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, $f(x)\geq f(\xi)$ for all $\xi\in M$. But $f\circ T=f$, so this means $f(Tx)\geq f(T\xi)$ for all $\xi\in M$. But $M$ is mapped to itself by the transformation  $T$, so this is the same as saying that $f(Tx)\geq f(\xi)$ 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 $f\colon M\to\mathbb{R}$, 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 ProofOfArithmeticgeometricMeansInequalityUsingLagrangeMultipliers 2013-03-22 15:25:53 2013-03-22 15:25:53 stevecheng (10074) stevecheng (10074) 16 stevecheng (10074) Example msc 49-00 msc 26B12 ArithmeticGeometricMeansInequality UsingJensensInequalityToProveTheArithmeticGeometricHarmonicMeansInequality