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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.

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.

J | > 1 and |αj+1 | ≤ 1, |αj+2 | ≤ 1, . . , |αn | ≤ 1. Prove that j i=1 |αi | ≤ |a0 |2 + |a1 |2 + · · · + |an |2 . ) 6. Prove that, for any real numbers x, y, z, 3(x2 − x + 1)(y 2 − y + 1)(z 2 − z + 1) ≥ (xyz)2 + xyz + 1. 7. (a) Prove that any polynomial P (x) such that P (x) ≥ 0 for all real x can be written as the sum of the squares of two polynomials. (b) Prove that the polynomial x2 (x2 − y 2 )(x2 − 1) + y 2 (y 2 − 1)(y 2 − x2 ) + (1 − x2 )(1 − y 2 ) is everywhere nonnegative, but cannot be written as the sum of squares of any number of polynomials.

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