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.

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, . . , 5, where xn+5 = xn for all n. Prove that x1 = · · · = x5 . 4. (USAMO 1979/3) Let x, y, z ≥ 0 with x + y + z = 1. Prove that 1 x3 + y 3 + z 3 + 6xyz ≥ . 4 34 5. (Taiwan, 1995) Let P (x) = 1 + a1 x + · · · + an−1 xn−1 + xn be a polynomial with complex coefficients.

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 , . . , xn ) is convex in a region if and only if the Hessian matrix Hij = ∂2 ∂xi ∂xj 30 is positive definite everywhere in the region. Note that the Hessian is symmetric because of the symmetry of mixed partials, so this statement makes sense. Proof. The function f is convex if and only if its restriction to each line is convex, and the second derivative along a line through x in the direction of y is (up to a scale factor) just Hy · y evaluated at x.