A is less than B by Kedlaya K.S.

By Kedlaya K.S.

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Xn ). ∂xi ∂xi Putting these conditions together with the constraint on g, one may be able to solve and thus put restrictions on the locations of the extrema. ) It is even more critical here than in the one-variable case that the Lagrange multiplier condition is a necessary one only for an interior extremum. Unless one can prove that the given function is convex, and thus that an interior extremum must be a global one, one must also check all boundary situations, which is far from easy to do when (as often happens) these extend to infinity in some direction.

BLAH. 5 1. (IMO 1968/2) Prove that for all real numbers x1 , x2 , y1 , y2 , z1 , z2 with x1 , x2 > 0 and x1 y1 > z12 , x2 y2 > z2 , the inequality 1 1 8 ≤ + 2 2 (x1 + x2 )(y1 + y2 ) − (z1 + z2 ) x1 y1 − z1 x2 y2 − z22 is satisfied, and determine when equality holds. (Yes, you really can apply the material of this section to the IMO! 6 Constrained extrema and Lagrange multipliers In the multivariable realm, a new phenomenon emerges that we did not have to consider in the one-dimensional case: sometimes we are asked to prove an inequality in the case where the variables satisfy some constraint.

Of course, we say f is concave if −f is convex. The analogue of the second derivative test for convexity is the Hessian criterion. A symmetric matrix M (that is, one with Mij = Mji for all i, j) is said to be positive definite if M x·x > 0 for all nonzero vectors x, or equivalently, if its eigenvalues are all real and positive. (The first condition implies the second because all eigenvalues of a symmetric matrix are real. ) Theorem 27 (Hessian test). A twice differentiable function f (x1 , . .

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