proof of arithmetic-geometric means inequality using Lagrange multipliers
with equality if and only if all the are equal.
To begin with, define by .
is almost a manifold (a differentiable surface), except that it has a boundary consisting of points with for some . Clearly attains the minimum zero on this boundary, and on the “interior” of , that is , is positive. Now is a manifold, and the maximum of on is evidently a local maximum on . Hence the Lagrange multiplier method may be applied.
We have the constraint11Although does not constrain that , this does not really matter — for those who are worried about such things, a rigorous way to escape this annoyance is to simply define to be zero whenever for some . Clearly the new cannot have a maximum at such points, so we may simply ignore the points with when solving the system. And for the same reason, the fact that fails to be differentiable at the boundary points is immaterial. under which we are to maximize .
from which it follows that ( is obviously inadmissible). Then substituting back, we get , which implies that the are all equal to , The value of at this point is .
Since we have obtained a unique solution to the Lagrange multiplier system, this unique solution must be the unique maximum point of on . So amongst all numbers whose arithmetic mean is , with equality occurring if and only if all the are equal to .
Proof of concavity
The question of whether the point obtained from solving Lagrange system is actually a maximum for 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 is a local maximum for , this time by showing that the Hessian is negative definite.
Actually, it turns out to be not much harder to show that is weakly concave everywhere on , not just on . A plausibility argument for this fact is that the graph of is concave, and since also involves an th root, it should be concave too. We will prove concavity of by showing that the Hessian of is negative semi-definite.
Since is a convex22 If 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 to will also be weakly concave; then we can conclude that the critical point on must be a global minimum on .
(The symbol is the Kronecker delta.) Thus the Hessian matrix is:
Now if we had actual numbers to substitute for the , 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 be a test vector. Negative semi-definiteness means for all . So let us write out:
We would be done if we can prove
Moreover, the case of equality in the Cauchy-Schwarz inequality (1) only occurs when is parallel to , i.e. for some scalar . Notice that is the normal vector to the tangent space of , so unless . This means, equality never holds in (1) if is restricted to the tangent space of . In turn, this means that is strictly concave on the convex set , and so is a strict global minimum of on .
Proof of inequality by symmetry argument
We have already calculated that is strictly concave on ; this was independent of the Lagrange multiplier method. Since is strictly concave on a closed convex set, it has a unique maximum there; call it .
Note that the argument in this last proof is extremely general — it takes the form: if is the unique maximum of , and is any symmetry of such that , then . 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|
|Date of creation||2013-03-22 15:25:53|
|Last modified on||2013-03-22 15:25:53|
|Last modified by||stevecheng (10074)|