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We have a pretty good understanding now of the relationship between extremization and the derivative, for functions in a single variable. However, most inequalities in nature deal with functions in more than one variable, so it behooves us to extend the formalism of calculus to this setting. Note that whatever we allow the domain of a function to be, the range of our “functions” will always be the real numbers. ) 29 The formalism becomes easier to set up in the language of linear algebra. e. a ball around) a point x, then we say the limit of f at x exists and equals L if for every > 0, there exists δ > 0 such that 0 < ||t|| < δ =⇒ |f (x + t) − L| < .

The bad news about this criterion is that determining whether a matrix is positive definite is not a priori an easy task: one cannot check M x · x ≥ 0 for every vector, so it seems one must compute all of the eigenvalues of M , which can be quite a headache. The good news is that there is a very nice criterion for positive definiteness of a symmetric matrix, due to Sylvester, that saves a lot of work. Theorem 28 (Sylvester’s criterion). An n × n symmetric matrix of real numbers is positive definite if and only if the determinant of the upper left k × k submatrix is positive for k = 1, .

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, . . , 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 ≥ .