A is less than B by Kedlaya K.S.

By Kedlaya K.S.

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Xn ) under the constraint g(x1 , . . , xn ) = c is the existence of λ such that ∂f ∂g (x1 , . . , xn ) = λ (x1 , . . , 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.

2 Additional problems Here is an additional collection of problems covering the entire range of techniques we have introduced, and one or two that you’ll have to discover for yourselves! 2 1. Let x, y, z > 0 with xyz = 1. Prove that x + y + z ≤ x2 + y 2 + z 2 . 2. The real numbers x1 , x2 , . . , xn belong to the interval [−1, 1] and the sum of their cubes is zero. Prove that their sum does not exceed n/3. 3. (IMO 1972/2) Let x1 , . . , x5 be positive reals such that (x2i+1 − xi+3 xi+5 )(x2i+2 − xi+3 xi+5 ) ≤ 0 for i = 1, .

N. Proof. By the M x · x definition, the upper left k × k submatrix of a positive definite matrix is positive definite, and by the eigenvalue definition, a positive definite matrix has positive determinant. Hence Sylvester’s criterion is indeed necessary for positive definiteness. We show the criterion is also sufficient by induction on n. 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.

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