A is less than B by Kedlaya K.S.

By Kedlaya K.S.

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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. 31 The Lagrange multiplier criterion for an interior local extremum of the function f (x 1 , . . , xn ) under the constraint g(x1 , . . , xn ) = c is the existence of λ such that ∂f ∂g (x1 , .

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.

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. So f is convex if and only if Hy · y > 0 for all nonzero y, that is, if H is positive definite.

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