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Then the Lagrange multiplier condition is that a = 2λx, b = 2λy, c = 2λz. The only points satisfying this condition plus the original constraint are ±√ 1 (a, b, c), a 2 + b 2 + c2 and these are indeed the minimum and maximum for f under the constraint, as you may verify geometrically. 1 Quick reference Here’s a handy reference guide to the techniques we’ve introduced. 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!

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. Suppose the roots of P (x) are α1 , α2 , . . , αn with |α1 | > 1, |α2 | > 1, . . , |αj | > 1 and |αj+1 | ≤ 1, |αj+2 | ≤ 1, . . , |αn | ≤ 1.

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 ) = 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.

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