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:


with equality if and only if all the xi are equal.

To begin with, define f:0n0 by f(x)=(x1xn)1/n.

Fix a>0 (the arithmetic meanMathworldPlanetmath), and consider the set M={xn:xi0 for all iixi=na}. This is clearly a closed and bounded set in n, and hence is compactPlanetmathPlanetmath. Therefore f on M has both a maximum and minimum.

M is almost a manifold (a differentiableMathworldPlanetmathPlanetmath surface), except that it has a boundary consisting of points with xi=0 for some i. Clearly f attains the minimum zero on this boundary, and on the “interior” of M, that is M={xn:xi>0 for all iixi=na}, f is positivePlanetmathPlanetmath. Now M is a manifold, and the maximum of f on M is evidently a local maximumMathworldPlanetmath on M. Hence the Lagrange multiplier method may be applied.

We have the constraint11Although g does not constrain that xi>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 xi0 for some i. Clearly the new f cannot have a maximum at such points, so we may simply ignore the points with xi0 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)=ixi-na under which we are to maximize f.

We compute:


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


from which it follows that λ=1/n (λ=-1/n is obviously inadmissible). Then substituting back, we get f(x)=xi, which implies that the xi 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)a amongst all numbers xi whose arithmetic mean is a, with equality occurring if and only if all the xi are equal to a.

The case a=0, which was not included in the above analysisMathworldPlanetmath, is trivial.

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 argumentMathworldPlanetmath. Another popular way to determine whether a given stationary point is a local maximum or minimum is by studying the HessianMathworldPlanetmath of the function at that point. For the sake of illustrating the relevant techniques, we will prove again that xi=a is a local maximum for f, this time by showing that the Hessian is negative definitePlanetmathPlanetmath.

Actually, it turns out to be not much harder to show that f is weakly concavePlanetmathPlanetmath everywhere on 0n, not just on M. A plausibility argument for this fact is that the graph of ttn is concave, and since f also involves an nth 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 restrictionPlanetmathPlanetmathPlanetmath of f to M will also be weakly concave; then we can conclude that the critical point xi=a on M must be a global minimum on M.

We begin by calculating second-order derivativesPlanetmathPlanetmath:

2fxjxi =xj(f(x)nxi)=1nxifxj+f(x)xj1nxi

(The symbol δij is the Kronecker delta.) Thus the Hessian matrix is:


Now if we had actual numbers to substitute for the xi, 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 algebraPlanetmathPlanetmath, and work directly from the definition.

Let v=(v1,,vn) be a test vector. Negative semi-definiteness means vD2f(x)vT0 for all v. So let us write out:

vD2f(x)vT =f(x)n2(i,jvixivjxj-nivi2xi2)

We would be done if we can prove

(iwi)2niwi2,wi=vixi. (1)

But look, this is just the Cauchy-Schwarz inequality, concerning the dot productMathworldPlanetmath of the vectors (w1,,wn) and (1,,1)! So D2f(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 (w1,,wn) is parallelMathworldPlanetmathPlanetmathPlanetmath to (1,,1), i.e. vi=μxi for some scalar μ. Notice that (1,,1) is the normal vector to the tangent spaceMathworldPlanetmath of M, so (1,,1)v=μixi0 unless μ=0. This means, equality never holds in (1) if v0 is restricted to the tangent space of M. In turn, this means that f is strictly concave on the convex set M, and so xi=a is a strict global minimum of f on M.

Proof of inequality by symmetry argument

Observe that the maximum point xi=a is pleasantly symmetricMathworldPlanetmathPlanetmathPlanetmathPlanetmath. We can show that the solution has to be symmetric, by exploiting the symmetryMathworldPlanetmath 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 ith and jth coordinatesMathworldPlanetmathPlanetmath of its argument.

By definition, f(x)f(ξ) for all ξM. But fT=f, so this means f(Tx)f(Tξ) for all ξM. But M is mapped to itself by the transformationMathworldPlanetmath T, so this is the same as saying that f(Tx)f(ξ) for all ξM. But the global maximum is unique, so Tx=x, that is, xi=xj.

Note that the argument in this last proof is extremely general — it takes the form: if x is the unique maximum of f:M, and T is any symmetry of M such that fT=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 inequalityMathworldPlanetmath: among all closed curves of a given arc lengthMathworldPlanetmath, the circle maximizes the area enclosed. However, Steiner did not realize that the assumptionPlanetmathPlanetmath 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